











^*f 












ADVANCED CALCULUS 



A TEXT UPON SELECT PARTS OF DIFFERENTIAL CAL- 
CULUS, DIFFERENTIAL EQUATIONS, INTEGRAL 
CALCULUS, THEORY OF FUNCTIONS, 
WITH NUMEROUS EXERCISES 



BY 



EDWIN BIDWELL WILSON, Ph.D. 

PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 



CHAPTERS I-X 



GINN AND COMPANY 

BOSTON • NEW YORK • CHICAGO ■ LONDON 



^ 



ENTERED AT STATIONERS' HALL 



COPYRIGHT, 1911, BY 
EDWIN BID WELL WILSON 



ALL RIGHTS RESERVED 
911.9 












G1NN AND COMPANY • PRO- 
PRIETORS • BOSTON • U.S.A. 



©■CLA295837 








/ 



//A 



PEEFACE 

It is probable that almost every teacher of advanced calculus feels the 
need of a text suited to present conditions and adaptable to his use. To 
write such a book is extremely difficult, for the attainments of students 
who enter a second course in calculus are different, their needs are not 
uniform, and the viewpoint of their teachers is no less varied. Yet in 
view of the cost of time and money involved in producing an Advanced 
Calculus, in proportion to the small number of students who will use it, 
it seems that few teachers can afford the luxury of having their own 
text ; and that it consequently devolves upon an author to take as un- 
selfish and unprejudiced a view of the subject as possible, and, so far as 
m him lies, to produce a book which shall have the maximum flexibility 
and adaptability. It was the recognition of this duty that has kept the 
present work in a perpetual state of growth and modification during 
five or six years of composition. Every attempt has been made to write 
in such a manner that the individual teacher may feel the minimum 
embarrassment in picking and choosing what seems to him best to meet 
the needs of any particular class. 

As the aim of the book is to be a working text or laboratory manual 
for classroom use rather than an artistic treatise on analysis, especial 
attention has been given to the preparation of numerous exercises which 
should range all the way from those which require nothing but substi- 
tution in certain formulas to those which embody important results 
withheld from the text for the purpose of leaving the student some 
vital bits of mathematics to develop. It has been fully recognized that 
for the student of mathematics the work on advanced calculus falls in 
a period of transition, — of adolescence, — in which he must grow from 
close reliance upon his book to a large reliance upon himself. More- 
over, as a course in advanced calculus is the ultima Thule of the 
mathematical voyages of most students of physics and engineering, it 
is appropriate that the text placed in the hands of those who seek that 
goal should by its method cultivate in them the attitude of courageous 



£~2- 



iv PREFACE 

explorers, and in its extent supply, not only their immediate needs, but 
much that may be useful for later reference and independent study. 

With the large necessities of the physicist and the growing require- 
ments of the engineer, it is inevitable that the great majority of our 
students of calculus should need to use their mathematics readily and 
vigorously rather than with hesitation and rigor. Hence, although due 
attention has been paid to modern questions of rigor, the chief desire 
has been to confirm and to extend the student's working knowledge of 
those great algorisms of mathematics which are naturally associated 
with the calculus. That the compositor should have set "vigor" where 
"rigor" was written, might appear more amusing were it not for the 
suggested antithesis that there may be many who set rigor where vigor 
should be. 

As I have had practically no assistance with either the manuscript 
or the proofs, I cannot expect that so large a work shall be free from 
errors ; I can only have faith that such errors as occur may not prove 
seriously troublesome. To spend upon this book so much time and 
energy which could have been reserved with keener pleasure for vari- 
ous fields of research would have been too great a sacrifice, had it not 
been for the hope that I might accomplish something which should be 
of material assistance in solving one of the most difficult problems of 
mathematical instruction, — that of advanced calculus. 

EDWIN BIDWELL WILSON 
Massachusetts Institute of Technology 



CONTENTS 



INTRODUCTORY REVIEW 



CHAPTER I 



REVIEW OF FUNDAMENTAL RULES 



SECTIOX 

1. On differentiation 
4. 



Logarithmic, exponential, and hyperbolic functions 
6. Geometric properties of the derivative 
8. Derivatives of higher order 

10. The indefinite integral . 

13. Aids to integration .... 

16. Definite integrals .... 



PAGE 

1 

4 
7 
11 
15 
18 
24 



CHAPTER II 

REVIEW OF FUNDAMENTAL THEORY 

18. Numbers and limits .... 

21. Theorems on limits and on sets of points 

23. Real functions of a real variable 

26. The derivative 

28. Summation and integration 



33 

37 
40 
45 
50 



PART I. DIFFERENTIAL CALCULUS 



CHAPTER III 
TAYLOR'S FORMULA AND ALLIED TOPICS 

31. Taylor's Formula .......... 55 

33. Indeterminate forms, infinitesimals, infinites . . . . .61 

36. Infinitesimal analysis ......... 68 

40. Some differential geometry ........ 78 

v 



VI 



CONTENTS 
CHAPTER IV 



PARTIAL DIFFERENTIATION 

SECTION 

43. Functions of two or more variables 
46. First partial derivatives . 
50. Derivatives of higher order 



EXPLICIT FUNCTIONS 



54. Taylor's Formula and applications 



PAGE 

87 

93 

102 

112 



CHAPTER V 

PARTIAL DIFFERENTIATION ; IMPLICIT FUNCTIONS 

56. The simplest case ; F (x, y) = 

59. More general cases of implicit functions 

62. Functional determinants or Jacobians 

65. Envelopes of curves and surfaces 

68. More differential geometry 



117 
122 
129 
135 
143 



CHAPTER VI 
COMPLEX NUMBERS AND VECTORS 



70. Operators and operations 

71. Complex numbers . 

73. Functions of a complex variable 

75. Vector sums and products 

77. Vector differentiation 



149 
153 
157 
163 
170 



PART II. DIFFERENTIAL EQUATIONS 
CHAPTER VII 

GENERAL INTRODUCTION TO DIFFERENTIAL EQUATIONS 



81. Some geometric problems 

83. Problems in mechanics and physics 

85. Lineal element and differential equation 

87. The higher derivatives ; analytic approximations 



179 
184 
191 
197 



CHAPTER VIII 

THE COMMONER ORDINARY DIFFERENTIAL EQUATIONS 

89. Integration by separating the variables ...... 203 

91. Integrating factors ......... 207 

95. Linear equations with constant coefficients . . . . . 214 

98. Simultaneous linear equations with constant coefficients . . 223 



CONTEXTS vii 
CHAPTER IX 
ADDITIONAL TYPES OF ORDINARY EQUATIONS 

SECTION PAGE 

100. Equations of the first order and higher degree .... 228 

102. Equations of higher order ........ 234 

104. Linear differential equations ....... 240 

107. The cylinder functions ......... 247 

CHAPTER X 

DIFFERENTIAL EQUATIONS IN MORE THAN TWO VARIABLES 

109. Total differential equations - 254 

111. Systems of simultaneous equations 260 

113. Introduction to partial differential equations .... 267 

116. Types of partial differential equations . . . . . . 273 



ADVANCED CALCULUS 

INTRODUCTORY REVIEW 

CHAPTER I 

REVIEW OF FUNDAMENTAL RULES 

1. On differentiation. If the function f(x) is interpreted as the 
curve y =f(x),* the quotient of the increments Ay and Ax of the 
dependent and independent variables measured from (x Q , y ) is 

y - ?/o = A y = */(*) = f( x o + Ax ) -f( x o) i (1) 

x — x Q Ax Ax Ax ^ ' 

and represents the slope of the secant through the points P (x Q , y) and 
P'(x -+- Ax, y + Ay) on the curve. The limit approached by the quo- 
tient Ay /Ax when P remains fixed and Ax = is the slope of the 
tangent to the curve at the point P. This limit, 

1 . m ^ =1 . m / fe± ^- Z (x 2 ) =/ , 

Ax = AX Ax = £±% 

is called the derivative of f(x) for the value x = x . As the derivative 
may be computed for different points of the curve, it is customary to 
speak of the derivative as itself a function of x and write 

lim Ay = Um f(x + *x)-f(x) = f 

^x = o Ax Ax = o Ax 

There are numerous notations for the derivative, for instance 

* Here and throughout the work, where figures are not given, the reader should draw 
graphs to illustrate the statements. Training in making one's own illustrations, whether 
graphical or analytic, is of great value. 

1 



2 INTKODUCTOBY EEYIEW 

The first five show distinctly that the independent variable is x, whereas 
the last three do not explicitly indicate the variable and should not be 
used unless there is no chance of a misunderstanding. 

2. The fundamental formulas of differential calculus are derived 
directly from the application of the definition (2) or (3) and from a 
few fundamental propositions in limits. First may be mentioned 

dz dz dii . , . 

Tx = i7j Tx ' where s = +<*> and y =f^- < 4 > 

(5) 



dx _ 
dy 


.df- 1 (y)_ 

dy 


1 
df(x) 

dx 


1 

dy' 

dx 




D (u ±v) = 


■ Du ± Dv, 




D(uv) = 


= uDv -f- vDu. 


K-)= 


vDu — vD 

v 2 


u m 


D(x n ) = 


= nx n ~ 1 . 



(6) 
(7) 

It may be recalled that (4), which is the rule for differentiating a function of a 
function, follows from the application of the theorem that the limit of a product is 

the product of the limits to the fractional identity — = ; whence 

Ax Ay Ax 



Az 
lim — = 


.. Az .. Ay .. , Az .. Ay 
lim lim — - = lim f lim — - 


\x = o Ax 


Ax~oAy Aa; = oAX Ay = Ay Aa; = oAX 



which is equivalent to (4). Similarly, if y =f(x) and if x, as the inverse function 
of y, be written x—f~ 1 (y) from analogy with ?/ = sinx and x = sin- 1 y, the 
relation (5) follows from the fact that Ax/ Ay and Ay /Ax are reciprocals. The next 
three result from the immediate application of the theorems concerning limits of 
sums, products, and quotients (§ 21). The rule for differentiating a power is derived 
in case n is integral by the application of the binomial theorem. 

Ay (x + Ax) n — x n , n (n — 1) „ . , . . n 

— = ^-^ = nx n ~ l + — i '-x n ~ 2 Ax + • • • + (Ax)"- 1 , 

Ax Ax 2 ! V ' - 

and the limit when Ax = is clearly nx" -1 . The result may be extended to rational 

values of the index n by writing n = — , y = x?, y f i — xp and by differentiating 

both sides of the equation and reducing. To prove that (7) still holds when n is 
irrational, it would be necessary to have a workable definition of irrational numbers 
and to develop the properties of such numbers in greater detail than seems wise at 
this point. The formula is therefore assumed in accordance with the 'principle of 
permanence of form (§ 178), just as formulas like a m a n = a m + n of the theory of 
exponents, which may readily be proved for rational bases and exponents, are 
assumed without proof to hold also for irrational bases and exponents. See, how- 
ever, §§ 18-25 and the exercises thereunder. 

* It is frequently better to regard the quotient as the product u ■ v- 1 and apply (6) . 
f For when sx = 0, then Ay = or Ay/ Ax could not approach a limit. 



FUNDAMENTAL RULES 3 

3. Second may be mentioned the formulas for the derivatives of the 
trigonometric and the inverse trigonometric functions. 



o 



D sin x = cos x, D cos x = — sin x, (8) 

D simr = sin(.>' + I-77-), D cos x = cos (x + ^ ir), - (8') 

X> tana = sec 2 x, D cot x = — esc 2 a?, . (9) 

D sec x = sec a? tan x, D esc ;/• = — esc x cot x, (10) 

D vers a; = sin x, -where vers x=l— cos x = 2 sin 2 \ x, (11) 

_ . +1 f + in quadrants I, IV, /H ON 

=: ^' 1-" " n - m ' ( } 

, +1 f — in quadrants I, II. /H ON 

DCOS-^= .- - > ^, M M TTT TV (13) 



Vi^7 2 l+" " m - IV 

i^l' Deot-^^--^ 

1 + x- 1 + .'' 



D tan" 1 x =-- = , D cot" 1 a- = - — — : 2 > (11) 



D sec -1 a? = 



jD eschar = 



~fc-*- f + in quadrants I, III. /H ~ N 

+ 1 f — in quadrants I. III. /H „ N 

wra' U« •■ ii. iv. (16) 

t. i ±1 f 4- in quadrants I, II. /H — 

■Dvers- 1 a-= . => *? « m 'J (1<) 

It may be recalled that to differentiate sin x the definition is applied. Then 

A sin x sin (x 4- Ax) — sin x sin Aa: 1 — cos Ax . 

= = cos x sm x. 

Ax Ax Ax Ax 

It now is merely a question of evaluating the two limits which thus arise, namely, 

.. sinAz .. 1— cosAz ' 

hm and hm (18) 

at=o Ax ax=o Ax 

From the properties of the circle it follows that these are respectively 1 and 0. 
Hence the derivative of sin a: is cosx. The derivative of cos a: may he found in 
like manner or from the identity cos x = sin (| it — x) . The results for all the other 
trigonometric functions are derived by expressing the functions in terms of sinz 
and cos a;. And to treat the inverse functions, it is sufficient to recall the general 
method in (5) . Thus 

if y = sin -1 a;, then &my = x. 



Differentiate both sides of the latter equation and note that cos?/ = ± VI — sin 2 ?/ 
= ± Vl — x 2 and the result for D sin -1 a: is immedi ate. T o ascertain which sign to 
use with the radical, it is sufficient to note that ± Vl — a.* 2 is cosy, which is positive 
when the angle y = sm- 1 x is in quadrants I and IV, negative in II and III. 
Similarly for the other inverse functions, 



INTRODUCTORY REVIEW 



EXERCISES 



1. Cany through the derivation of (7) when n = p/q, and review the proofs of 
typical formulas selected from the list (5)-(17). Note that the formulas are often 
given as D x ii n = nu n - 1 D x u, D x sin u = cos u D x u, • • • , and may be derived in this 
form directly from the definition (3). 

2. Derive the two limits necessary for the differentiation of since. 

3. Draw graphs of the inverse trigonometric functions and label the portions 
of the curves which correspond to quadrants I, II, III, IV. Verify the sign in 
(12)-(17) from the slope of the curves. 

4. Find D tanx and D cotx by applying the definition (3) directly. 

5. Find J) sin x by the identity sin u — sin v = 2 cos sin 

2 2 

6. Find D tan- 1 ^ by the identity tan- 1 ^ — tan- 1 v = tan- 1 and (3). 

1 + uv 

7. Differentiate the following expressions : 



(a) esc 2 x — cot 2 x, (/3) i tan 3 x — tan x + x, (7) x cos -1 x — Vl — x 2 , 

(5) sec- 1 , (e) sin- 1 — , (f) x Va 2 — x 2 + a 2 sin- 1 - , 

-Vl-x 2 Vl + x 2 a 

2 ax •■ • , x 



(77) a vers -1 V2 ax — x 2 , (#) cot- 1 ■ 2 tan- 1 

a x 2 — a 2 a 

What trigonometric identities are suggested by the answers for the following : 

(a) sec 2 x, (5) 1 , (e) _* (0)0? 

Vl - x 2 1 + a 2 

8. In B. 0. Peirce's " Short Table of Integrals " (revised edition) differentiate the 
right-hand members to confirm the formulas : Nos. 31, 45-47, 91-97, 125, 127-128, 
131-135, 161-163, 214-216, 220, 260-269, 294-298, 300, 380-381, 386-394. 

9. If x is measured in degrees, what is D sinx ? 

4. The logarithmic, exponential, and hyperbolic functions. The 

next set of formulas to be cited are 

Dlog e x = l, Dlog a x = 1 -^, (19) 

De x = e x , Da x = a x log e a. t (20) 

It may be recalled that the procedure for differentiating the logarithm is 

X 

Alog a x log rt (x + Ax) — log a x 1 , x + Ax 1. L , Ax\±x 

— -^ = -2-^ = — loga — = - l0g a 1 + — ) • 

Ax Ax Ax x x \ x ) 

* The student should keep on file his solutions of at least the important exercises ; 
many subsequent exercises and considerable portions of the text depend on previous 
exercises. 

t As is customary, the subscript e will hereafter be omitted and the symbol log will 
denote the logarithm to the base e ; any base other than e must be specially designated 
as such. This observation is particularly necessary with reference to the common base 
10 used in computation. 



FUNDAMENTAL RULES 5 

If now x/Ax be set equal to h, the problem becomes that of evaluating 

lim (l + -Y' = e = 2.71828 • • • ,* log 10 e = 0.434294 . . . ; (21) 

and hence if e be chosen as the base of the system, D log x takes the simple form 
1/x. The exponential functions e x and cFmay be regarded as the inverse functions 
of log x and log«x in deducing (21). Further it should be noted that it is frequently 
useful to take the logarithm of an expression before differentiating. This is known 
as logarithmic differentiation and is used for products and complicated powers and 
roots. Thus 

if y — x x , then log y = x log x, 

and - y' = 1 + log x or y' = x x (1 + log x) . 

It is the expression y'/y which is called the logarithmic derivative of y. An especially 
noteworthy property of the function y = Ce x is that the function and its derivative 
are equal, y' — y ; and more generally the function y = Ce kx is proportional to its 
derivative, y' — ky. 

5. The hyperbolic functions are the hyperbolic sine and cosine, 

pK p % pX _I .y X 

sinh x = ■ — — — j cosh x = - — — — ; (22) 

and the related functions tanh a;, cotha:, seeha?, eschar, derived from 
them by the same ratios as those by which the corresponding trigono- 
metric functions are derived from since and cos a'. From these defini- 
tions in terms of exponentials follow the formulas : 

cosh 2 x — sinh 2 as = 1, tanh 2 a' + sech. 2 as = 1, (23) 

sinh (x ± y) = sinh x cosh y + cosh x sinh y, (24) 

cosh (x + y) = cosh x cosh y + sinh x sinh y, * (25) 



cosh 



x cosh x -f- 1 . . x , Icosh x — 1 
2 = + aJ 2 ' slnh 2 =± >J 2~ 



(26) 



D sinh a? = cosh x, D cosh a; = sinh x, (27) 

£> tanh x = sech 2 a?, D cotli sc = — csch' 2 a-, (28) 

D sech x = — sech a? tanh a;, Z> csch x = — csch a: coth a:. (29) 

The inverse functions are expressible in terms of logarithms. Thus 

e 2 ■' — 1 

y = sinh -1 x. x = sinh y = — — , 

* > * 2 e y 

6^ — 2^^ — 1=0, e y = a; + Va- 2 +l. 

* The treatment of this limit is far from complete in the majority of texts. Reference 
for a careful presentation may, however, be made to Granville's "Calculus," pp. 31-34, 
and Osgood's " Calculus," pp. 78~82. See also Ex. 1, (/3), in § 165 below. 



6 INTRODUCTORY REVIEW 

Here only the positive sign is available, for e y is never negative. Hence 

sinh- 1 x = log (x + Vrr 2 + l), any x, (30) 

cosh -1 x = log (x ± Va? 2 — l), x 2 > 1, (31) 

1 1+sc 

tanh- 1 ^ = - log — — , x 2 <l, (32) 

,ii 1 1 x + 1 

coth- 1 jc = - log — - , cc 2 > 1, (33) 



secli -1 x = log ( - ± -^ 



i-11, *•<!, 



csch" 1 oj = log /- + ^- + 1 j ? any x, (35) 

D sinh" 1 x = , + , D cosh" 1 x = , ~ , (36) 

D tanh- 1 ^ = 5 = Z> coth- 1 ^ = 5 , (37) 

1— x 2 1—x 2 v y 

D sech" 1 x = — — > D csch" 1 jc = ~ ■ (38) 

xVl-x* x^l+x 2 



EXERCISES 

1. Show by logarithmic differentiation that 

•^ / v /w' v' w' \ . 

D (uvw • • •) = 1 1 [■•■•) (wow • • •), 

v ' \u v w / v " 

and hence derive the rule : To differentiate a product differentiate each factor 
alone and add all the results thus obtained. 

2. Sketch the graphs of the hyperbolic functions, interpret the graphs as those 
of the inverse functions, and verify the range of values assigned to x in (30)-(35). 

3. Prove sundry of formulas (23)-(29) from the definitions (22). 

4. Prove sundry of (30)-(38), checking the signs with care. In cases where 
double signs remain, state when each applies. Note that in (31) and (34) the 
double sign may be placed before the log for the reason that the two expressions 
are reciprocals. 

5. Derive a formula for sinhw ± sinhv by applying (24) ; find a formula for 
tanh \x analogous to the trigonometric formula tan \x = sinx/(l + cosx). 

6. The gudermannian. The function = gdx, defined by the relations 

sinh x = tan 0, <f> = gd x = tan- 1 sinh x, — |7r<0< + |7r, 
is called the gudermannian of x. Prove the set of formulas : 
cosh x = sec 0, tanh x = sin 0, csch x = cot 0, etc. ; 
D gd x = sech x, x = gd- 1 = log tan (*- + \ tt), D gd- 1 = sec 0. 

7. Substitute the functions of in Ex. 6 for their hyperbolic equivalents in 
(23), (26), (27), and reduce to simple known trigonometric formulas. 



FUNDAMENTAL EULES 7 

8. Differentiate the following expressions : 

(a) (x + l) 2 (x + 2)-3(x + 3)-4, (/3) x^s^ ( 7 ) \ ogx ( x + i), 

(5) x + log cos(x — i 7r), (e) 2 tan- 1 ^, (f) x — tanhx, 

, , • , ,„ o, ,•* e f{X (a. sin mx — m cos mx) 
(t?) xtanli-ix + |log(l-x 2 ), (0) — * ^ ^ '-• 

9. Check sundry formulas of Peirce's " Table," pp. 1-61, 81-82. 

6. Geometric properties of the derivative. As the quotient (1) and 
its limit (2) give the slope of a secant and of the tangent, it appears 
from graphical considerations that when the derivative is positive the 
function is increasing with x, but decreasing when the derivative is 
negative.* Hence to determine the regions in which a function is in- 
creasing or decreasing, one may find the derivative and determine the 
regions in which it is positive or negative. 

One must, however, be careful not to apply this rule too blindly ; for in so 
simple a case as f(x) = log x it is seen that f'{x) = 1/x is positive when x > and 
negative when x < 0, and yet log x has no graph when x < and is not considered 
as decreasing. Thus the formal derivative may be real when the function is not 
real, and it is therefore best to make a rough sketch of the function to corroborate 
the evidence furnished by the examination of f'{x). 

If x is a value of x such that immediately t upon one side of x = x Q 
the function f(x) is increasing whereas immediately upon the other 
side it is decreasing, the ordinate y o =f(x ) will be a maximum or 
minimum or f(x) will become positively or negatively infinite -at x Q . 
If the case where f(x) becomes infinite be ruled out, one may say that 
the function will have a minimum or maximum at x Q according as the 
derivative changes from negative to positive or from positive to negative 
when x, moving in the ptositive direction, j^asses through the value x . 
Hence the usual rule for determining maxima and minima is to find 
the roots off'(x) = 0. 

This rule, again, must not be applied blindly. For first, /"(x) may vanish where 
there is no maximum or minimum as in the case y = x 3 at x = where the deriva- 
tive does not change sign ; or second, /'(x) may change sign by becoming infinite 
as in the case y = x» at x = where the curve has a vertical cusp, point down, and 
a minimum ; or third, the function /(x) may be restricted to a given range of values 
a == x == o for x and then the values /(a) and/(6). of the function at the ends of the 
interval will in general be maxima or minima without implying that the deriva- 
tive vanish. Thus although the derivative is highly useful in determining maxima 
and minima, it should not be trusted to the complete exclusion of the corroborative 
evidence furnished by a rough sketch of the curve y =f{x). 

* The construction of illustrative figures is again left to the reader. 

f The word "immediately" is necessary because the maxima or minima may be 
merely relative ; in the case of several maxima and minima in an interval, some of 
the maxima may actually be less than some of the minima. 



INTKODUCTOKY EEVIEW 



7. The derivative may be used to express the equations of the tangent 
and normal, the values of the subtangent and subnormal, and so on. 

Equation of tangent, y — y = y' (% 
Equation of normal, (y — y ) y[-\-(x- 



-*»)> 


(39) 


x ) = 0, 


(40) 


'= subnormal = y n y[, 


(41) 


\-yJy'v ete - 


(42) 




TM = subtangent = yjy\, MN 
OT = ^--intercept of tangent == x 

The derivation of these results is sufficiently evi- 
dent from the figure. It may be noted that the 
subtangent, subnormal, etc., are numerical values 
for a given point of the curve but may be regarded 
as functions of x like the derivative. 
In geometrical and physical problems it is frequently necessary to 
apply the definition of the derivative to finding the derivative of an 
unknown function. For instance if A denote the 
area under a curve and measured from a fixed 
ordinate to a variable ordinate, A is surely a func- 
tion A (x) of the abscissa x of the variable ordinate. 
If the curve is rising, as in the figure, then 

MPQ'M' < AA< MQP'M', or yAx < AA <(jj + Ay) Ax. 
Divide by Ax and take the limit when Ax = 0. There results 




lim y 

Ax = 



Hence 



lim — — g 

Aa: = Ax 

lim — = 

Ax = Ax 



lim (y + Ay). 



Ax = 

dA 
dx 



(43) 



Rollers Theorem and the Theorem of the Mean are two important 
theorems on derivatives which will be treated in the next chapter but 
may here be stated as evident from their geometric interpretation. 
Eolle's Theorem states that : If a function has a derivative at every 




Y 


V 


^? 


' 




A 


*r 


B 







1 






Fig. 2 



Fig. 3 



point of an interval and if the function vanishes at the ends of the in- 
terval, then there is at least one point within the interval at which the 
derivative vanishes. This is illustrated in Fig. 1, in which there are 
two such points. The Theorem of the Mean states that : If a function 



FUNDAMENTAL EULES 9 

has a derivative at each point of an interval, there is at least one pjoint 
in the interval such that the tangent to the curve y=f(x) is parallel to 
the chord of the interval. This is illustrated in Fig. 2 in which there 
is only one such point. 

Again care must be exercised. In Fig. 3 the function vanishes at A and B but 
there is no point at which the slope of the tangent is zero. This is not an excep- 
tion or contradiction to Rolle's Theorem for the reason that the function does not 
satisfy the conditions of the theorem. In fact at the point P, although there is a 
tangent to the curve, there is no derivative ; the quotient (1) formed for the point P 
becomes negatively infinite as Ax = from one side, positively infinite as Ax = 
from the other side, and therefore does not approach a definite limit as is required 
in the definition of a derivative. The hypothesis of the theorem is not satisfied and 
there is no reason that the conclusion should hold. 

EXERCISES 

1. Determine the regions in which the following functions are increasing or 
decreasing, sketch the graphs, and find the maxima and minima : 

( a ) i x 3 - x 2 + 2, (/3) (x + 1)£ (x - 5) 3 , (7) log (x 2 - 4), 

(8) (x - 2)Vx-l, (e) - (x + 2)Vl2-x 2 , (f) x 3 + ax + b. 



2. The ellipse is r = Vx 2 + y 2 = e (d + x) referred to an origin at the focus. 
Find the maxima and minima of the focal radius r, and state why D x r = does 
not give the solutions while B^r = does [the polar form of the ellipse being 
r — k{l — 6COS0)- 1 ]. 

3. Take the ellipse a s x 2 /a? + y 2 /b 2 = 1 and discuss the maxima and minima of 
the central radius r = Vx 2 + y 2 . Why does D x r = give half the result when r is 
expressed as a function of x, and why will B x r = give the whole result when 
x = a cos A, y = b sin \ and the ellipse is thus expressed in terms of the eccentric 
angle ? 

4xlf y = P (x) is a polynomial in x such that the equation P (x) = has multiple 
roots, show that P'{x) = for each multiple root. What more complete relationship 
can be stated and proved ? 

5. Show that the triple relation 27 b 2 + 4 a 3 = determines completely the nature 
of the roots of x 3 + ax + b = 0, and state what corresponds to each possibility. 

6. Define the angle 6 between two intersecting curves. Show that 

tan# = [/'(x ) - 0'(x o )] + [1 +/ / (x )/(x )] 
if y =f(x) and y = g (x) cut at the point (x , y ). 

7. Find the subnormal and subtangent of the three curves 

(a) y 2 = 4px, (/3) x 2 = 4py, (7) x 2 + y 2 = a 2 . 

8. The pedal curve. The locus of the foot of the perpendicular dropped from 
a fixed point to a variable tangent of a given curve is called the pedal of the given 
curve with respect to the given point. Show that if the fixed point is the origin, 
the pedal of y =f{x) may be obtained by eliminating x , y , y'o from the equations 

y-y Q = y'o (» - x ) , yy' Q + x = o, y = f(x ) , y = f (x ) . 



10 INTEODTTCTOEY EEYIEW 

Find the pedal (a) of the hyperbola with respect to the center and (/3) of the 
parabola with respect to the vertex and (7) the focus. Show (5) that the pedal of 
the parabola with respect to any point is a cubic. 

9. If the curve y =f(x) be revolved about the x-axis and if V(x) denote the 
volume of revolution thus generated when measured from a fixed plane perpen- 
dicular to the axis out to a variable plane perpendicular to the axis, show that 

D X V = TO/ 2 . 

10. More generally if A (x) denote the area of the section cut from a solid by 
a plane perpendicular to the x-axis, show that D X V — A (x). 

11. If A (0) denote the sectorial area of a plane curve r =f{<t>) and be measured 
from a fixed radius to a variable radius, show that D<f>A = \r 2 . 

12. If /o, h, p are the density, height, pressure in a vertical column of air, show 
that dp/dh = — p. If p = kp, show p = Ce~ kh . 

13. Draw a graph to illustrate an apparent exception to the Theorem of the 
Mean analogous to the apparent exception to Rolle's Theorem, and discuss. 

14. Show that the analytic statement of the Theorem of the Mean for/(x) is 
that a value x — £ intermediate to a and b may be found such that 

f(b) -f(a) = /'((j (b - a), a < £ < b. 

15. Show that the semiaxis^of an ellipse ;is a mean proportional between the 
x-intercept of the tangent and the abscissa of the point of contact. 

16. Find the values of the length of the tangent (a) from the point of tangency 
to the x-axis, (j8) to the ?/-axis, (7) the total length intercepted between the axes. 
Consider the same problems for the normal (figure on page 8) . 

17. Find the angle of intersection of (a) y 2 = 2 rax and x 2 + y 2 = a 2 , 

/ox 9 a t 8aS /\ x2 , y 2 1 for 0<X<6 

(A x* = iay and 2/ = ^^ « ^— - t + ^3^ = 1 and6 <,<«. 

18. A constant length is laid off along the normal to a parabola. Find the locus. 

19. The length of the tangent to x% + y* = a? intercepted by the axes is constant. 

20. The triangle formed by the asymptotes and any tangent to a hyperbola has 
constant area. 

21. Find the length PT of the tangent to x = Vc 2 — y 2 + c sech- 1 (y/c). 

22. Find the greatest right cylinder inscribed in a given right cone. 

23. Find the cylinder of greatest lateral surface inscribed in a sphere. 

24. From a given circular sheet of metal cut out a sector that will form a cone 
(without base) of maximum volume. 

25. Join two points A, B in the same side of a line to a point P of the line in 
such a way that the distance PA + PB shall be least. 

26. Obtain the formula for the distance from a point to a line as the minimum 
distance. 

27. Test for maximum or minimum, {a) If f(x) vanishes at the ends of an inter- 
val and is positive within the interval and if f'(x) = has only one root in the 
interval, that root indicates a maximum. Prove this by Rolle's Theorem. Apply 
it in Exs. 22-24. (/3) If f(x) becomes indefinitely great at the ends of an interval 
and/'(x) = has only one root in the interval, that root indicates a minimum. 



FUNDAMENTAL EULES 11 

Prove by Rolle's Theorem, and apply in Exs. 25-26. These rules or various modi- 
fications of them generally suffice in practical problems to distinguish between 
maxima and minima without examining either the changes in sign of the first 
derivative or the sign of the second derivative ; for generally there is only one 
root of /'(#) = in the region considered. 

28. Show that x- 1 sinx from £=r0tox = |7r steadily decreases from 1 to 2/tt. 

1 l£ 2 

29. If 0<x < 1, show (a) 0< x — log (1 + x) <-x 2 , (/3) -? <x-log(l + x). 

2t 1 -f- X 

1 A X 2 

30. If > x > - 1, show that -x 2 < x - log (1 + x) < -* — - • 

2 1 + x 

8. Derivatives of higher order. The derivative of the derivative 
(regarded as itself a function of x) is the second derivative, and so on 
to the nth. derivative. Customary notations are : 

/"(*) = ^ = § = *>:/= »ly = y" = &/= &v, 

The ?ith derivative of the sum or difference is the sum or difference of 
the nth. derivatives. Eor the nt\i derivative of the product there is a 
special formula known as Leibniz's Theorem. It is 

D n (uv) = D n u • v + nD n - 1 icDv + —^— — "- D n -' 2 uD 2 v-\ \-uD n v. (44) 

This result may be written in symbolic form as 

Leibniz's Theorem D n (uv) = (Du + Dv) n , (44') 

where it is to be understood that in expanding (Du -\- Dv) n the term 
(Du) k is to be replaced by D k u and (Dm) by D°u = u. In other words 
the powers refer to repeated differentiations. 

A proof of (44) by induction will be found in § 27. The following proof is 
interesting on account of its ingenuity. Note first that from 

D (uv) = uDv + vDu, D 2 (uv) = D (uDv) + -D (vDu), 

and so on, it appears that D 2 (uv) consists of a sum of terms, in each of which there 
are two differentiations, with numerical coefficients independent of u and v. In like 
manner it is clear that 

D*(uv) = C D"u ■ v + C^-iuDv + • • • + C n ^iDuD»-^v + C n uD"v 

is a sum of terms, in each of which there are n differentiations, with coefficients C 
independent of u and v. To determine the C's any suitable functions u and ■», say, 

u = e x , v = e ax , uv = e^ 1 + a ) a: , D k ~e ax = a k e ax , 

may be substituted. If the substitution be made and e( 1+a ) x be canceled, 

e-(i+«)xDn ( , (r) = (1+ a y = c o + C x a + ... + C n ^a^-X + C n a», 

and hence the C's are the coefficients in the binomial expansion of (1 + a) n . 



12 INTRODUCTORY REVIEW 

Formula (4) for the derivative of a function of a function may be 
extended to higher derivatives by repeated application. More generally 
any desired change of variable may be made by the repeated use of (4) 
and (5). For if x and y be expressed in terms of known functions 
of new variables u and v, it is always possible to obtain the deriva- 
tives D x y, D^y, • • • in terms of D u v, D*v, • ■ -, and thus any expression 
F(x, y, y', y", • • •) may be changed into an equivalent expression 
&(u, v, v', v", • • •) in the new variables. In each case that arises the 
transformations should be carried out by repeated application of (4) 
and (5) rather than by substitution in any general formulas. 

The following typical cases are illustrative of the method of change of variable. 
Suppose only the dependent variable y is to be changed to z defined as y =f{z) . Then 

d 2 y _ d /dy\ _ d /dz dy\ _ d 2 z dy dz I d dy\ 

dx 2 dx \dx) dx \dx dz) dx 2 dz dx \dx dz) 

_ d 2 z dy dz Id dy dz\ _ d 2 z dy /dz\ 2 d 2 y 

dx 2 dz dx \dz dz dx) dx 2 dz \dx) dz 2 

As the derivatives of y =f{z) are known, the derivative d 2 y/dx 2 has been expressed 
in terms of z and derivatives of z with respect to x. The third derivative would be 
found by repeating the process. If the problem were to change the independent 
variable x to z, defined by x =f{z), 

dy _dy dz _ dy /dxX- 1 d 2 y _ d Ydy /dx\- 1_ | 
dx dz dx dz\dz) dx 2 dx\_dz \dz) J 



d 2 y _ d 2 y dz 
d^~~dz^dx 



(dx\~ 1 _ dy fdx\- 2 dzd 2 x_ Vd^y dx _ <Px dyl ^_ /dxY 
\dz) dz \dz) dx~dz 2 ~ Ldz 2 dz ~ dz 2 dz] ' \dz) 



The change is thus made as far as derivatives of the second order are concerned. If 
the change of both dependent and independent variables was to be made, the work 
would be similar. Particularly useful changes are to find the derivatives of y by x 
when y and x are expressed parametrically as functions of t, or when both are ex- 
pressed in terms of new variables r, <f> as x = r cos 0, y = r sin <p. For these cases 
see the exercises. 

9. The concavity of a curve y —f(x) is given by the table : 
if /"(x ) > 0, the curve is concave up at x = x oK 

if /" (x Q ) < 0, the curve is concave down at x = x , 

if /"(a; ) = 0, an inflection point at x = x . (?) 

Hence the criterion for distinguishing between maxima and minima : 
if /' (x ) = and /" (a? ) > 0, a minimum at x = x , 

if /' (x ) = and /" (x Q ) <0, a maximum at x = x , 

if /' (x ) = and /" (x Q ) = 0, neither max. nor min. (?) 



FUNDAMENTAL EULES 13 

The question points are necessary in the third line because the state- 
ments are not always true unless f'"(x ) =f= (see Ex. 7 under § 39). 

It may be recalled that the reason that the curve is concave up in case/"^) > 
is because the derivative f'[x) is then an increasing function in the neighborhood 
of x = x ; whereas if f"{x ) < 0, the derivative f'(x) is a decreasing function and 
the curve is convex up. It should be noted that concave up is not the same as 
concave toward the x-axis, except when the curve is below the axis. With regard 
to the use of the second derivative as a criterion for distinguishing between maxima 
and minima, it should be stated that in practical examples the criterion is of rela- 
tively small value. It is usually shorter to discuss the change of sign of f\x) directly, 
— and indeed in most cases either a rough graph of f[x) or the physical conditions 
of the problem which calls for the determination of a maximum or minimum will 
immediately serve to distinguish between them (see Ex. 27 above). 

The second derivative is fundamental in dynamics. By definition the 
average velocity v of a particle is the ratio of the space traversed to the 
time consumed, v = s/t. The actual velocity v at anj T time is the limit 
of this ratio when the interval of time is diminished and approaches 
zero as its limit. Thus 

As As ds .._ 

v — — and v = lim — = — • (4o) 

At At=0^t clt v } 

In like manner if a particle describes a straight line, say the cc-axis, the 
average acceleration f is the ratio of the increment of velocity to the 
increment of time, and the actual acceleration f at any time is the limit 
of this ratio as M = 0. Thus 

- Av Av dv d 2 x .... 

/=- and /= h ? _ = - = _. (46) 

By JYeivton's Second Law of Motion, the force acting on the particle is 
equal to the rate of change of momentum with the time, momentum 
being defined as the product of the mass and velocity. Thus 

„ d(mv) dv . d 2 x /4 _ N 

dt dt J dt 2 

where it has been assumed in differentiating that the mass is constant, 
as is usually the case. Hence (47) appears as the fundamental equa- 
tion for rectilinear motion (see also §§ 79, 84). It may be noted that 

dv __d_/l A dT 



where T = \ mv 2 denotes by definition the kinetic energy of the particle. 
For comments see Ex. 6 following. 



14 ESTTBODUCTOKY EEVIEW 

EXERCISES 

1. State and prove the extension of Leibniz's Theorem to products of three or 
more factors. Write out the square and cube of a trinomial. 

2. Write, by Leibniz's Theorem, the second and third derivatives : 

(a) e x sin x, (/3) cosh x cos x, (7) x 2 e x log x. 

3. Write the nth derivatives of the following functions, of which the last three 
should first be simplified by division or separation into partial fractions. 



(a) Vx + 1, OS) log (ax + &), (7) {x 2 + 1) (x + 1)~ ; 

(5) cosax, (e) e^sinx, , (f) (1 — x)/(l + x), 

/ \ I m\ xS ± x + X / \ / ^ + A 2 



4. If 2/ and x are each functions of t, show that 
dx d 2 y dy d 2 x 
d 2 y dt dt 2 dt dt 2 x'y" — y'x' 



dx 2 /dx" 



(dxy x 6 

\dt) 

d s y _ x'jx'y'" — y'x'") — 3-x"(x'y" — y'x") 
dx 3 x' 5 

5. Find the inflection points of the curve x = 4 — 2 sin 0, y = A — 2 cos <f>. 

6. Prove (47'). Hence infer that the force which is the time-derivative of the 
momentum mv by (47) is also the space-derivative of the kinetic energy. 

7. If A denote the area under a curve, as in (43), find dA/dO for the curves 

(a) y = a (1 — cos #), x = a(6 — sin 0), (/3) x = a cos 0, y = b sin 6. 

8. Make the indicated change of variable in the following equations: 

(g) ^ + _j^^+ V =0 , x = tans. 4ns . ** + „ = 0. 

v ' cix 2 1 + x 2 dx (1 + X 2 ) 2 ' <*Z 2 

{/J) (l-x^-I^Yl-x^+^O, , = *, x = sin M . 

\_dx 2 y \dx/ J dx $2 V 

Arts. 1-1 = 0. 

du 2 

9. Transformation to polar coordinates. Suppose thatx = rcos0, y=r sin 0. Then 

dx dr . dy dr . t 

— = — cos — r sin 0, — = — sin + r cos 0, 

d0 d0 dcp d<p 

dy , d 2 y _r 2 +2(D <}) r) 2 -rD 2 r 

— ■ and ■ — — • 

dx dx 2 (cos D^r — r sin 0) 3 

10. Generalize formula (5) for the differentiation of an inverse function. Find 
d 2 x/dy 2 and d s x/dy s . Note that these may also be found from Ex. 4. 

11. A point describes a circle with constant speed. Find the velocity and 
acceleration of the projection of the point on any fixed diameter. 

12. Prove — -=2uv z + 4v i [ — ) — v 5 ( — ifx = -,y = uv. 

dx 2 \duj du 2 \duj v 



FUNDAMENTAL RULES 15 

10. The indefinite integral. To integrate a function f(x) is to find 
a function F(x) the derivative of which is f(x). The integral F(x) is 
not uniquely determined by the integrand f(x) ; for any two functions 
which differ merely by an additive constant have the same derivative. 
In giving formulas for integration the constant may be omitted and 
understood ; but in applications of integration to actual problems it 
should always be inserted and must usually be determined to lit the 
requirements of special conditions imposed upon the problem and 
known as the initial conditions. 

It must not be thought that the constant of integration always appears added to the 
function F (x) . It may be combined with F (x) so as to be somewhat disguised. Thus 

logs, logx + C, logCx, \og(x/C) 

are all integrals of 1/x, and all except the first have the constant of integration C, 
although only in the second does it appear as formally additive. To illustrate the 
determination of the constant by initial conditions, consider the problem of finding 
the area under the curve y = cosx. By (43) 

D X A = y = cos x and hence A = sin x + C. 

If the area is to be measured from the ordinate x = 0, then A = when x = 0, and 
by direct substitution it is seen that C = 0. Hence A = sinx. But if the area be 
measured from x=— |tt, then A = when x= — \ir and C =1. Hence A = 1 + sinx. 
In fact the area under a curve is not definite until the ordinate from which it is 
measured is specified, and the constant is needed to allow the integral to fit this 
initial condition. 

11. The fundamental formulas of integration are as follows : 



f l x = lo S x, 


I x n = x n if n 4= — 1, 

J nJ r 1 


(48) 


J e x = e x , 


I a x = « x /log a, 


(49) 


J sin x = — cos x, 


I cos x = sin x, 


(50) 


I"""- -'" ""' 


j cot x = log sin x, 


(51) 


/■—-■ • 


Jcsc^-cota, 


(52) 


I tan x sec x = sec x, 


| cot X CSC X = — CSC X, 


(53) 



with formulas similar to (50)— (53) for the hyperbolic functions. Also 

/- j = tan _1 cc or — cot -1 ;r, / :, = tanh _1 x or coth -1 sc, (54) 
1 + 03 J 1 — X 



16 INTRODUCTORY REVIEW 

/l C ±1 

— =sin~ 1 a; or — cos _1 £c, I — , — + sinh -1 a;, (55) 

Vl-^ 2 J Vl + cc 2 ~ * \- ) 

/l /» +1 
— = sec _1 a5 or — csc _1 a;, j = =F sech _1 2c, (56) 

/+ 1 r ± i 
~ = ± cosh -1 a?. / ~ ==F csch _1 x, (57) 

/ . = = vers -1 x, I sec x = gd _1 x = log tan ( — + - ) • (58) 

V 2 x — x' 2 J \4 2/ 

For the integrals expressed in terms of the inverse hyperbolic functions, the 
logarithmic equivalents are sometimes preferable. This is not the case, however, 
in the many instances in which the problem calls for immediate solution with 

regard to x. Thus if y = / (1 + £ 2 ) - i = sinh- 1 x + C, then x = sinh (y—C), and the 

solution is effected and may be translated int o expo nentials. This is not so easily 
accomplished from the form y = log (x + Vl + x 2 ) + C. For this reason and 
because the inverse hyperbolic functions are briefer and offer striking analogies 
with the inverse trigonometric functions, it has been thought better to use them 
in the text and allow the reader to make the necessary substitutions from the table 
(30) -(35) in case the logarithmic form is desired. 

12. In addition to these special integrals, which are consequences 
of the corresponding formulas for differentiation, there are the general 
rules of integration which arise from (4) and (6). 

J dy dx J dx ^ ' 

J (ii + v — to) = I u + I v — I w, (60) 

tcv = I uv' -f- / u'v. (61) 

Of these rules the second needs no comment and the third will be treated later. 
Especial attention should be given to the first. For instance suppose it were re- 
quired to integrate 2 logx/x. This does not fall under any of the given types ; but 

2. d(loffx) 2 dlogx dz dy 

- log X = v & ' ^— = 

x ' d log x dx dy dx 

Here (logx) 2 takes the place of z and logx takes the place of y. The integral is 
therefore (logx) 2 as may be verified by differentiation. In general, it may be 
possible to see that a given integrand is separable into two factors, of which one 
is integrable when considered as a function of some function of x, while the other 
is the derivative of that function. Then (59) applies. Other examples are : 

C e smx cosa ^ ftan- 1 x/(l + x 2 ), Tx 2 sin(x 3 ). 



FUNDAMENTAL EULES IT 

In the first, z = ev is integrable and as y — since, y' = cosx ; in the second, z — y is 
integrable and as y = tan- 1 ^, y / = (1 + x 2 )- 1 ; in the third z — sin y is integrable 
and as y = x 3 , ?/' = 3x 2 . The results are 

e sina: , | (tan- 1 x) 2 , — i cos (x 3 ) . 

This method of integration at sight covers such a large percentage of the cases 
that arise in geometry and physics that it must be thoroughly mastered.* 



EXERCISES 

1. Verify the fundamental integrals (48)-(58) and give the hyperbolic analogues 
of (50)-(53). 

2. Tabulate the integrals here expressed in terms of inverse hyperbolic func- 
tions by means of the corresponding logarithmic equivalents. 

3. Write the integrals of the following integrands at sight : 

(a) sin ax, (/3) cot(ax + &), (7) tanh3x, 

1 1 ,.,1 



(5) 
(v) 



a 2 + x 2 
1 

X loo- X 



(p) 


cot (ax 
1 


+ &), 


Vx 2 - 
1 


a 2 


(tf) 


e x 
X~ 2 ' 





(k) x 3 Vax 2 + 6, (X) tan x sec 2 x, 

. , (x- 1 — l) 5 . . tanh-ix 

(") 15 ' (°) 



(p) a 1 + sinx cosx, (<t) 



1-x' 2 
sinx 

Vcosx 



\w 


V2 ax — x 2 


(0 


X 


x 2 + a 2 ' 


(/*) 


tan x log sin x, 


(»■) 


2 + logx 

> 

X 


(T) 


1 


Vl — x 2 sin- 1 x 



4. Integrate after making appropriate changes such as sin 2 x = | — \ cos 2x 
or sec 2 x = 1 + tan 2 a;, division of denominator into numerator, resolution of the 
product Ox trigonometric functions into a sum, completing the square, and so on. 

(a) cos 2 2x, (fi) sin 4 x. (7) tan 4 x, 

(8) X , (e) !£±i-, (D 1 - 8i11 * , 

v x 2 + 3x + 25 v x + 2 v/ versx 

x + 3 e 2 *+e* , , 1 



4x 2 -5x + l e^ + 1 V2ax + x 2 

( /c ) sin 5 x cos 2 x + 1, (X) smb. mx sinh nx, (/x) cos x cos 2 x cos 3 x,. 

(v) sec 5 xtanx — V2x, (0) — , (77-) — 



x 2 + ax + 6 (ax™ + b)p 

* The use of differentials (§ 35) is perhaps more familiar than the use of derivatives. 

»W-/|*-/|f*-/|*-«l*(»)]- 



Then 



I - log xdx = I 2 log a: a" log x = (log x)'- 



The use of this notation is left optional with the reader ; it has some advantages and 
some disadvantages. The essential thing is to keep clearly in mind the fact that the 
problem is to be inspected with a view to detecting the function which will differentiate 
into the given integrand. 



18 INTEODUCTOEY REVIEW 

5. How are the following types integrated ? 

(a) sin w x cos n x, m or n odd, or ra and n even, 
(/3) tan w x or cot n x when n is an integer, 

(7) sec w x or csc n x when n is even, 

(8) tan w x sec n x or cot w x csc w x, n even. 

6. Explain the alternative forms in (54)-(56) with all detail possible. 

7. Find (a) the area under the parabola y 2 = 4px from x = to x = a ; also 
(/3) the corresponding volume of revolution. Find (7) the total volume of an ellip- 
soid of revolution (see Ex. 9, p. 10). 

8. Show that the area under y — sin mx sin nx or y = cos mx cos nx from x = 
to x = 7T is zero if m and w are unequal integers but | ir if they are equal. 

9. Eind the sectorial area of r = a tan0 between the radii = and $ = \ie. 

10. Eind the area of the (a) lemniscate r 2 = a 2 cos 2 and (£) cardioidr=l — cos0. 

11. By Ex. 10, p. 10, find the volumes of these solids. Be careful to choose the 
parallel planes so that A (x) may be found easily. 

(a) The part cut off from a right circular cylinder by a plane through a diameter 
of one base and tangent to the other. Ans. 2/3 it of the whole volume. 

(j8) How much is cut off from a right circular cylinder by a plane tangent to its 
lower base and inclined at an angle 6 to the plane of the base ? 

(7) A circle of radius b < a is revolved, about a line in its plane at a distance a 
from its center, to generate a ring. The volume of the ring is 2 7r%/A 

(5) The axes of two equal cylinders of revolution of radius r intersect at right 
angles. The volume common to the cylinders is 16r 3 /3. 

12. If the cross section of a solid isA(x) = a x 3 + a x x 2 + a 2 x + a 3 , a cubic in x, 
the volume of the solid between two parallel planes is \h(B + 4 M + B') where h 
is the altitude and B and B / are the bases and M is the middle section. 

x + c 



J 1 + x 2 



13. Show that = tan- 1 



1 — ex 



13. Aids to integration. The majority of cases of integration which 
arise in simple applications of calculus may be treated by the method 
of § 12. Of the remaining cases a large number cannot be integrated 
at all in terms of the functions which have been treated np to this 

point. Thus it is impossible to express / . in terms 

1 J V(l-* 2 )(l-aV) 

of elementary functions. One of the chief reasons for introducing a 
variety of new functions in higher analysis is to have means for effect- 
ing the integrations called for by important applications. The dis- 
cussion of this matter cannot be taken up here. The problem of 
integration from an elementary point of view calls for the tabula- 
tion of some devices which will accomplish the integration for a 



FUNDAMENTAL EXILES 19 

wide variety of integrands integrable in terms of elementary functions. 
The devices which will be treated are : 

Integration by parts, Eesolution into partial fractions, 

Various substitutions, Eeference to tables of integrals. 
Integration by parts is an application of (61) when written as 

Cuv' = uv- fu'v. (61') 

That is, it may happen that the integrand can be written as the product uv' of two 
factors, where v' is integrable and where u'v is also integrable. Then wo' is integrable. 
For instance, logx is not integrated by the fundamental formulas ; but 



I log x = I log x -l = x log x — / x/x = 



X lOgX 



Here log x is taken as u and 1 as r/, so that v is x, u' is 1/x, and u'v = 1 is immedi- 
ately integrable. This method applies to the inverse trigonometric and hyperbolic 
functions. Another example is 



| x sin x = — x cos x + / cos x = sin x 



Here if x = u and sin x = v', both v' and u'v = — cosx are integrable. If the choice 
sin x=u and x=v' had been made, v' would have been integrable but u'v = \ x 2 cos x 
would have been less simple to integrate than the original integrand. Hence in 
applying integration by parts it is necessary to look ahead far enough to see that 
both v' and u'v are integrable, or at any rate that v' is integrable and the integral 
of u'v is simpler than the original integral.* 

Frequently integration by parts has to be applied several times in succession. Thus 

Cx-e x = x 2 e x — f 2 xe x if u = x 2 , v' = e x , 

= x 2 e x —2\xe x - f V if 'it. = x, v' = e x , 

= x 2 e x — 2 xe x + 2 e x . 

Sometimes it may be applied in such a way as to lead back to the given integral 
and thus afford an equation from which that integral can be obtained by solution. 
For example, 

/ e* cos x = e* cos x + / e x sin x if u = cos x, v' = e x , 

= e*cos* + |Vsi„*-jVcos*] if M = sinx,„=e*, 

= ^(cosx + sinx) — / e x cosx. 
Hence j e^cosx = \ e x (cosx + since). 

* The method of differentials may again be introduced if desired. 



20 LXTKODUCTOBY REVIEW 

14. For the integration of a rational fraction f(x)/F(x) where /and Fare poly- 
nomials in x, the fraction is first resolved into partial fractions. This is accom- 
plished as follows. First if / is not of lower degree than F, divide F into / until the 
remainder is of lower degree than F. The fraction f/F is thus resolved into the 
sum of a polynomial (the quotient) and a fraction (the remainder divided by F) 
of which the numerator is of lower degree than the denominator. As the polyno- 
mial is integrable, it is merely necessary to consider fractions f/F where / is of 
lower degree than F. Next it is a fundamental theorem of algebra that a poly- 
nomial F may be resolved into linear and quadratic factors 

F(x) = k (x — a) a (x — 6)0 (x — c)v -. . . (x 2 + mx + n)/* (x 2 +px + q)v... <) 

where a, 6, c, • • • are the real roots of the equation F(x)=0 and are of the respec- 
tive multiplicities a, /3, 7, • • • , and where the quadratic factors when set equal to 
zero give the pairs of conjugate imaginary roots of F =1 0, the multiplicities of the 
imaginary roots being fi, v, • • ■ • . It is then a further theorem of algebra that the 
fraction f/F may be written as 

f(x) A'A„ Aa B, Be 

F(x) x — a (x — a) 2 (x — a) a x — b (x — 6)0 

M x x + N x M 2 x + N 2 M^x + Jy 

x 2 + mx + n (x 2 + mx + n) 2 (x 2 + mx + n)n 

where there is for each irreducible factor of fa term corresponding to the highest 
power to which that factor occurs in F and also a term corresponding to every 
lesser power. The coefficients A, 73, • • •, M, TV, • • • may be obtained by clearing 
of fractions and equating coefficients of like powers of x, and solving the equations ; 
or they may be obtained by clearing of fractions, substituting for x as many dif- 
ferent values as the degree of F, and solving the resulting equations. 

When f/F has thus been resolved into partial fractions, the problem has been 
reduced to the integration of each fraction, and this does not present serious 
difficulty. The following two examples will illustrate the method of resolution 
into partial fractions and of integration. Let it be required to integrate 

f *±2 __ and f 2 * 3 + 6 

J x (x - 1) (x - 2) (x 2 + x + 1) J (x - l) 2 (x - 3) 3 

The first fraction is expansible into partial fractions in the form 

x 2 + l A B C Dx + E 

= - + 7 + 7 + 






x (x — 1) (x — 2) (x 2 + x + 1) x x — 1 x — 2 x 2 + x + 1 

Hence x 2 + 1 = A(x - 1) (x - 2) (x 2 + x + 1) + Bx{x - 2) (x 2 + x + 1) 
+ Cx(x - 1) (x 2 + x + 1) + (Dx + E)x(x - 1) (x - 2). 

Kather than multiply out and equate coefficients, let 0, 1, 2, — 1, — 2 be substi- 
tuted. Then 

1 = 24, 2 =-3 3, 5 = 14(7, D -E = 1/21, E- 2D = 1/7, 

x 2 + l r 1 C 2 f 5 f4x+5 



f x z + i - C — C 4- C - c 

J x(x-lWx-2Wx 2 + x+l) ~ J 2x"~J 3(x-l) Jl4(x-2) J 



(£_l)( x _2)(x 2 + x+l) J 2x J 3(x-l) J 14 (x- 2) J21(x 2 + x + l) 

12 5 2 22x+l 
- logx — - log(x — 1) + — log(x — 2) log(x 2 + x + 1) tan- 1 — — - . 

2 to 3 &v ' 14 &v ' 21 oV 7V3 V3 



FUNDAMENTAL EULES 21 

In the second case the form to be assumed for the expansion is 

2 x 3 + 6 A B C B E 

+ - - + - —,+ 



(x - l) 2 (x - 3) 3 x - 1 (x - l) 2 (x - 3) (x- 3) 2 (x - 3) 3 
2 x 3 + 6 = A (x - 1) (x - 3) 3 + £ (x - 3) 3 + C(x- l) 2 (x - 3) 2 
+ B (x - l) 2 (x — 3) + E (x - l) 2 . 
The substitution of 1. 3, 0, 2, 4 gives the equations 

8 =-8 5, 60 = 4D, 9J. + 3C-D + 12 = 0, 
J._O + D + 6 = 0, A + 3 C + 3D = 0. 
The solutions are — 9/4, — 1, + 9/4, — 3/2, 15, and the integral becomes 



/ 



9 log(x-l) + -^- + ?log(x-3) 



(x-l) 2 (x-3) 3 4 ov ' x-1 4 

3 15 



2 (x - 3) 2 (x - 3) 2 

The importance of the fact that the method of partial fractions shows that any 
rational fraction may be integrated and, moreover, that the integral may at most con- 
sist of a rational part plus the logarithm of a rational fraction plus the inverse 
tangent of a rational fraction should not be overlooked. Taken with the method 
of substitution it establishes very wide categories of integrands which are inte- 
grable in terms of elementary functions, and effects their integration even though 
by a somewhat laborious method. 

15. The method of substitution depends on the identity 

f f(x) = f /[> (V)] ^ if x = (i/), (59') 

v x */ y ay 

which is allied to (59). To show that the integral on the right with respect to y 

is the integral of f(x) with respect to x it is merely necessary to show that its 

derivative with respect to x is f(x). By definition of integration, 

sX /[ * W] S= /[ *« ] S-i= /[ * W] . 

by (4). The identity is therefore proved. The method of integration by substitu- 
tion is in fact seen to be merely such a systematization of the method based on 
(59) and set forth in § 12 as will make it practicable for more complicated problems. 
Again, differentials may be used if preferred. 

Let B denote a rational function. To effect the integration of 

/ sinx B (sin 2 x, cosx), let cos x = y, then / — B (1 — ?/ 2 , y) ; 
| cosx E(cos 2 x, sinx), let sin x = y, then JB(l — y 2 ,y); 

/ B = / E(tana!) ' let tanx=y < *»*}.£&•• 

/iUsins.co**), let tan| = „, then j-B^Jfc, ^J)^- 

The last substitution renders any rational function of sin x and cos x rational in 
the variable y ; it should not be used, however, if the previous ones are applicable 
— it is almost certain to give a more difficult final rational fraction to integrate. 



22 INTRODUCTORY REVIEW 

A large number of geometric problems give integrands which are rational in x 
and in some one of the radicals Va 2 -f x 2 , Va 2 — x 2 , Vx 2 — a* 2 . These may be con- 
verted into trigonometric or hyperbolic integrands by the following substitutions : 



j R (x, Va 2 — x 2 ) x = a sin ?/, j R(a sin y, a cos ?/) a cos y ; 
| x = a tan 2/, j R(a tan ?/, a sec y) a sec 2 2/ 
\ x = a sinh ?/, I R(a sinh y, a cosh ?/) a cosh ?/ ; 

x — a sec y, j R(a sec ?/, a tan y) a sec ?/ tan y 
J y 

x = a cosh ?/, j R(a cosh z/, a sinh ?/) a sinh y. 



Cr (x, Va 2 + x 2 ) 



f P (x, Vx 2 - a 2 ) 



2/ 

It frequently turns out that the integrals on the right are easily obtained by 
methods already given ; otherwise they can be treated by the substitutions above. 

In addition to these substitutions there are a large number of others which are 
applied under specific conditions. Many of them will be found among the exer- 
cises. Moreover, it frequently happens that an integrand, which does not come 
under any of the standard types for which substitutions are indicated, is none the 
less integrable by some substitution which the form of the integrand will suggest. 

Tables of integrals, giving the integrals of a large number of integrands, have 
been constructed by using various methods of integration. B. O. Peirce's ff Short 
Table of Integrals " may be cited. If the particular integrand which is desired does 
not occur in the Table, it may be possible to devise some substitution which will 
reduce it to a tabulated form. In the Table are also given a large number of 
reduction formulas (for the most part deduced by means of integration by parts) 
which accomplish the successive simplification of integrands which could perhaps 
be treated by other methods, but only with an excessive amount of labor. Several 
of these reduction formulas are cited among the exercises. Although the Table is 
useful in performing integrations and indeed makes it to a large extent unneces- 
sary to learn the various methods of integration, the exercises immediately below, 
which are constructed for the purpose of illustrating methods of integration, should 
be done without the aid of a Table. 

EXERCISES 

1. Integrate the following by parts : 

(a) Ixcoshx, (j3) Jtan-ix, (7) jx'Mogx, 

? _. fsin-ix . , r xe x r 1 

2. If P(x) is a polynomial and P'(x), P"(x), • • • its derivatives, show 

(a) fP (x) e™ = - e«* \p (x) - - P'(x) + - a P"(x) 1 , 

J a \_ a a 2 J 

(/3) Cp (x) cos ax = -sin ax P (x) P"(x) + — P iv (x) 

J a \_ a 2 a^ J 

+ - cos ax \- T'(x) - - P'"(x) + — P v (x) 1 , 

a \_a a s - a 5 J 

and (7) derive a similar result for the integrand P (x) sin ax. 



(a) je ax smbx 



FUNDAMENTAL EULES 23 

3. By successive integration by parts and subsequent solution, show 
e^ (a sin bx — b cos bx) 

t"-~ Sill UJO = 

()8) fe a;,; cos6x = 

J a 2 + 6 2 

(7 ) l xe 2x cos x = 2V e ' 2 x [5 x (sin x + 2 cos x) — 4 sin x — 3 cos x] . 

4. Prove by integration by parts the reduction formulas 

sin" 1 + 1 x cos n _1 x n — 1 



a 2 + 6 2 
e^ (6 sin bx + a cos 6x) 



(a) I sm m x cos n x = 1 | sm m x cos w ~ 2 x, 

ra + n m + n J 

u n ni -ixsec w x m 

m + n — 1 m + 



,„. /* tai^'-ixsec^x m — 1 /\ 

(j8) I tan m xsec n x = / tan w - 2 xsec w x, 

J m + n — 1 m + ?i — 1 «/ 



(7) 



r l_ - I r ; +<2»-8) f i 1, 

J (x 2 + a 2 )" 2 (n - 1) a 2 L (x 2 + a 2 )' 1 - 1 ' J (x 2 + a 2 )" -* J 

/* x m x m + 1 m+lr x m 

^ J (logx) re ~~ ~ (}i-l)(logx)»-i n- lJ (logx)"- 1 ' 

5. Integrate by decomposition into partial fractions : 

K) J (x + 2) 2 (x + 1) ' J 2x 5 + x 3 K) Jx(l + x 2 ) 2 ' 

6. Integrate by trigonometric or hyperbolic substitution : 

(a) CVa 2 - x 2 , (/3) fVx?^~a?, (7) CVa 2 + x 2 , 

(.)/—!—. w^i wf^ 

J (a-x 2 )l ^ x J xi 

7. Find the areas of these curves and their volumes of revolution : 

(a) xf + ?/l = at, (/3) ahj 2 = a 2 x± - x 6 , (7) (^f + (f V = ] 

8. Integrate by converting to a rational algebraic fraction: 
sin3x .„, /• cos3x , ■ /• sin2x 



( ) c sm x (/3) r cos x ( ) c 

J a 2 cos 2 x + b 2 sin 2 x «/ a 2 cos 2 x + 6 2 sin 2 x J 

/n /* 1 C( 

a + 6 cos x ' «/ a + b cos x + c sin x «/ 1 + si 



a 2 cos 2 x + 6 2 sin 2 x 
cosx 

+ sinx 



9 . Show that I B (x, Va + 6x + ex 2 ) may be treated by trigonometric substitu- 
tion ; distinguish between b 2 — 4tac> 0. 

10. Show that CbIx. \ ) is made rational by y n = ' Hence infer 

J \ \cx + d/ ex + d 

that fit (x, V(x — a) (x — £)) is rationalized by y 2 = ■ This accomplishes 

J x — (X 

the integration of B (x, Vet + bx + ex 2 ) when the roots of a + foe + ex 2 = are 

real, that is, when b 2 — 4 ac > 0. 



24 INTKODUCTOBY EEVIEW 

1 1 . Show that / R x, ( J , ( ) , ■ • • , where the exponents m, n, 

J L \cx + ay \cx + d) J 

ax -1- b 
• • • are rational, is rationalized by y k = if k is so chosen that km, kn, • • • are 

ex + d 
integers. 

12. Show that I (a + by)pyi may be rationalized if p or g or p + g is an integer. 

By setting x n — y show that J x m (a + &x«)p may be reduced to the above type and 

. . -, i , m + 1 m 4- 1 

hence is intestable when or p or ]- p is integral. 

n n 

13. If the roots of a + 6x 4- ex 2 = are imaginary, J E (x, Va -f bx 4- ex 2 ) may 
be rationalized by y =_ Va + bx + ex 2 ^ x Vc. 

14 . Integrate the following . 

(") / -7==' (/3) / -—=, (7) / tt= — ?=' 

J Vx-1 J l+Vx J Vl + x-Vl + x 



X 3 


Vx-1 


e 2x 


Ve" + l' 


1 



(x — d) Va + 6x + ex 2 



(.)/-__=, («) r . , ■ (»r 

J -vV + 1 J V(l - x 2 ) 3 J 



x(l + x 2 )t J' x J VI -x 3 

15. In view of Ex. 12 discuss the integrability of : 

x™ flet x = a?/ 2 , 



sin m x cos n x, let sinx=V?/, (/3) I — -4 



Vax — x 2 L or Vax — x 2 = xy. 

16. Apply the reduction formulas, Table, p. 66, to show that the final integral for 

/x m . r 1 r x r 1 
______ i S i — _____ or j — ______ or j _______ 
Vl - x 2 J Vl - x 2 J Vl - x 2 J x Vl - x 2 

according as m is even or odd and positive or odd and negative. 

17. Prove sundry of the formulas of Peirce's Table. 



18. Sho w that if R (x, Va 2 — x 2 ) contains x only to odd powers, the substitu- 
tion z = Va 2 — x 2 will rationalize the expression. Use Exs. 1 (f) and 6 (e) to 
compare the labor of this algebraic substitution with that of the trigonometric or 
hyperbolic. 

16. Definite integrals. If an interval from x = a to x = b be divided 
into n successive intervals Ax 1} Ax 2 , • • ■, Ax n and the value /(£•) of a 
function f(x) be computed from some point £ in each interval Ax { and 
be multiplied by Ax { , then the limit of the sum 

im [/(£) _*_! +/(&) Ax, + • • • + /&) ^] = £/(») dx < ( 62 ) 



« = 00 



FUNDAMENTAL RULES 



25 



when each interval becomes infinitely short and their number n be- 
comes infinite, is known as the definite integral of f(x) from a to b, and 
is designated as indicated. If y=f(x) be graphed, the sum will be 
represented by the area under 
a broken line, and it is clear 
that the limit of the sum, that 
is, the integral, will be repre- 
sented by the area under the 
curve y =f(x) and between 
the ordinates x = a and x = b. 
Thus the definite integral, de- 
fined arithmetically by (62), 
may be connected with a geo- 
metric concept which can serve to suggest properties of the integral 
much as the interpretation of the derivative as the slope of the tan- 
gent served as a useful geometric representation of the arithmetical 
definition (2). 

Eor instance, if a, b, c are successive values of x, then 




£ /(*) <** + X /(*) (h = L a*) dx 



(63) 



is the equivalent of the fact that the area from a to c is equal to the 
sum of the areas from a to b and b to c. Again, if \x be considered 
positive when x moves from a to b, it must be considered negative 
when x moves from b to a and hence from (62) 



$J(x)dx = -fj{x)dx. 



(64) 



Finally, if M be the maximum of f(x) in the interval, the area under 
the curve will be less than that under the line y = M through the 
highest point of the curve ; and if m be the minimum of fix), the 
area under the curve is greater than that under y = m. Hence 



(b - a) <£f(x) dx < M(b - a). 



(65) 



There is, then, some intermediate value m < /x < M such that the inte- 
gral is equal to fx(b — a); and if the line y = fx cuts the curve in a 
point whose abscissa is £ intermediate between a and b, then 

£f(x) dx = M,(b- «) = (b - a)f(i). (65') 

This is the fundamental Theorem of the Mean for definite integrals. 



26 INTKODUCTOKY EEVIEW 

The definition (62) may be applied directly to the evaluation of the definite in- 
tegrals of the simplest functions. Consider first 1/x and let a, b be positive with a 
less than 6. Let the interval from a to 5 be divided into n intervals Axj which are 
in geometrical progression in the ratio r so that Xi = a, x 2 = ar, • • •, x n +i = ar n 

and Axi = a(r — 1), Ax 2 = ar (r — 1), Ax 3 = ar 2 (r — 1), • • •, Ax n = ar n ~ l (r — 1) ; 

whence b—a = Axi + Ax 2 + ••_■ + Ax n — a (r n — 1) and r n = b/a. 

Choose the points & in the intervals Axj as the initial points of the intervals. Then 

^1 + ^_ 2 + . . . + ^ = a ( r ~ 1 ) + ar{r-l) + . . . + ar»-i(r-l) = ,__ ^ 
£i |« l« a a*" ar* 1 - 1 

But r = V b/a or n = log (6/a) -^- log r. 

„ Axi Ax 2 Ax n , b r — 1 . 6 /i 

Hence — - H + 1 = n (r — 1) = log - • = log 



f i & In a log r a log (1 + h) 

Now if n becomes infinite, r approaches 1, and h approaches 0. But the limit of 
log (1 + h)/h as h = is by definition the derivative of log (1 + x) when x = and 
is 1. Hence 



J a X w=oo L £l £2 In J 



log - = log b — log a. 
a 



As another illustration let it be required to evaluate the integral of cos 2 x from 
to | 7r. Here let the intervals Ax 4 - be equal and their number odd. Choose the £'s 
as the initial points of their intervals. The sum of which the limit is desired is 

a- = cos 2 • Ax + cos 2 Ax • Ax + cos 2 2 Ax • Ax + • ■ ■ 

+ cos 2 (n — 2) Ax ■ Ax + cos 2 (n — 1) Ax • Ax. 

But nAx = 1 7r, and (n — 1) Ax = \ it — Ax, (n — 2) Ax = \ it — 2 Ax, • • •, 

and cos (| -k — y) = sin y and sin 2 y + cos 2 y = l. 



Hence a = Ax 

— Ax 



cos 2 + cos 2 Ax + cos 2 2 Ax + ■ • • + sin 2 2 Ax + sin 2 Ax] 



Hence / " cos 2 xdx = lim [± n Ax + % Ax] = lim (i tt + \ Ax) = \ it. 

Jo Ax = Ax = 

Indications for finding the integrals of other functions are given in the exercises. 

It should be noticed that the variable x which appears in the expression of the 
definite integral really has nothing to do with the value of the integral but merely 
serves as a symbol useful in forming the sum in (62). What is of importance is 
the function /and the limits a, b of the interval over which the integral is taken. 



f b f(x)dx= f b f(t)clt= f h f(y)dy= f /(*) 



(U . 



The variable in the integrand disappears in the integration and leaves the value of 
the integral as a function of the limits a and b alone. 



FUNDAMENTAL EXILES 27 

17. If the lower limit of the integral be fixed, the value 



i 



b 

f(x) dx = ® (b) 



of the integral is a function of the upper limit regarded as variable. 
To find the derivative $'(#), form the quotient (2), 

Xb + Ab r* b 

f(x)dx- \ f(x)dx 

Ab Ab 

By applying (63) and (65'), this takes the simpler form 

Jf>b + A6 
f(x) dx 

Ab Ab -&b' f{i)Ab ' 



where £ is intermediate between b and b -f- Ab. Let Ab = 0. Then £ 
approaches a and/(£) approaches /(a). Hence 

*' (J)= ^ ff( x ) dx =/(»)■ ( 66 ) 

If preferred, the variable b may be written as x, and 

*(*)= J /(*)<&, *'(*) = ^jf /(*)<& =/<*)• (66') 

This equation will establish the relation between the definite integral 
and the indefinite integral. For by definition, the indefinite integral 
F(x) of /(a) is any function such that F'(x) equals f(x). As &'(x) =f(x) 
it follows that ~ x 

J f(x)dx = F(x)+C. (67) 

Hence except for an additive constant, the indefinite integral of f is 
the definite integral of / from a fixed lower limit to a variable upper 
limit. As the definite integral vanishes when the upper limit coincides 
with the lower, the constant C is — F(a) and 

b 

f(x) dx = F(b)- F(a). (67') 



X 



Hence, the definite integral of f{x) from a to b is the difference between 
the values of any indefinite integral F(x) taken for the iqiper and lower 
limits of the definite integral ; and if the indefinite integral of f is 
known, the definite integral may be obtained without applying the 
definition (62) to / 



28 INTRODUCTORY REVIEW 

The great importance of definite integrals to geometry and physics 
lies in that fact that many quantities connected with geometric figures 
or physical bodies way be expressed simply for small portions of the 
figures or bodies and may then be obtained as the sum of those quanti- 
ties taken over all the small portions, or rather, as the limit of that sum 
when the portions become smaller and smaller. Thus the area under a 
curve cannot in the first instance be evaluated ; but if only that portion 
of the curve which lies over a small interval Ax be considered and the 
rectangle corresponding to the ordinate /(£) be drawn, it is clear that 
the area of the rectangle is /(£) Ax, that the area of all the rectangles is 
the sum %f(£) Ax taken from a to b, that when the intervals Ax approach 
zero the limit of their sum is the area under the curve ; and hence that 
area may be written as the definite integral of f(x) from a to b* 

In like manner consider the mags of a rod of variable density and suppose the 
rod to lie along the cc-axis so that the density may he taken as a function of x. 
In any small length Ax of the rod the density is nearly constant and the mass of 
that part is approximately equal" to the product pAx of the density p(x) at the 
initial point of that part times the length Ax of the part. In fact it is clear that 
the mass will be intermediate between the products mAx and MAx, where m and 
M are the minimum and maximum densities in the interval Ax. In other words 
the mass of the section Ax will be exactly equal to p (£) Ax where £ is some value of 
x in the interval Ax. The mass of the whole rod is therefore the sum 2/o(£)Ax 
taken from one end of the rod to the other, and if the intervals be allowed to 
approach zero, the mass may be written as the integral of p(x) from one end of 
the rod to the other.! 

Another problem that may be treated by these methods is that of finding the 
total pressure on a vertical area submerged in a liquid, say, in water. Let w be the 
weight of a column of water of cross section 1 sq. unit and 
of height 1 unit. (If the unit is a foot, w = 62.5 lb.) At a 
bOQ^\ point h units below the surface of the water the pressure is 

wh and upon a small area near that depth the pressure is 
approximately whA if A be the area. The pressure on the 
area A is exactly equal to w%A if £ is some depth interme- 
diate between that of the top and that of the bottom of 
the area. Now let the finite area be ruled into strips of height Ah. Consider the 
product wh b (h) Ah where b(h) =f(h) is the breadth of the area at the depth h. This 

* The £'s may evidently be so chosen that the finite sum 2/(£)Aa: is exactly equal to 
the area under the curve ; hut still it is necessary to let the intervals approach zero and 
thus replace the sum by an integral because the values of £ which make the sum equal 
to the area are unknown. 

t This and similar problems, here treated by using the Theorem of the Mean for 
integrals, may be treated from the point of view of differentiation as in § 7 or from that 
of Duhamel's or Osgood's Theorem as in §§ 34, 35. It should be needless to state that in 
any particular problem some one of the three methods is likely to be somewhat preferable 
to either of the others. The reason for laying such emphasis upon the Theorem of the 
Mean here and in the exercises below is that the theorem is in itself very important and 
needs to be thoroughly mastered. 



Ah 



FUNDAMENTAL EULES 29 

is approximately the pressure on the strip as it is the pressure at the top of the strip 
multiplied by the approximate area of the strip. Then w£b(£)Ah, where £ is some 
value between h and h + Ah, is the actual pressure on the strip. (It is sufficient to 
write the pressure as approximately whb(h)Ah and not trouble with the £.). The 
total pressure is then 2w>£&(£) Ah or better the limit of that sum. Then 

P = lim Vw£&($)dft = f whb(h)dh, 

where a is the depth of the top of the area and b that of the bottom. To evaluate 
the pressure it is merely necessary to find the breadth b as a function of h and 
integrate. 



1. 



EXERCISES 

b 



Jr> O pU 

kf(x) dx — k I f(x) dx. 
a J a 



Xb pb p b 

(u ± v) dx = / udx ± j vdx. 
v a «/ a 



Xb pb pb 

\p (x) dx < I f(x) dx < I (x) 
J a J a 



dx. 



4. Suppose that the minimum and maximum of the quotient Q(x) =f(x)/(p(x) 

of two functions in the interval from a to b are m and M, and let (x) be positive 

so that 

fix) 
m < Q(x) = -^-^ < M and m<p (x) <f(x)< M<p (x) 
0(x) 

are true relations. Show by Exs. 3 and 1 that 

f b f(x)dx f b f(x)dx 

m<^- b <M and ^ = fi =Q^) = iM 

f <p(x)dx f cp(x)dx ^^ 

where £ is some value of x between a and b. 

5. If m and M are the minimum and maximum of f(x) between a and b and if 
(x) is always positive in the interval, show that 

Xb pb pb 

(x) dx < I f(x) (x) dx < M I (x) dx 

Xb pb pb 

f (x) (x) dx = fj. I (p(x)dx =/(£) I (p(x)dx. 
. j. J a J a 

Note that the integrals of [M — f{x)] 0(x) and [/(#) — ni] <$> (x) are positive and 
apply Ex. 2. 

6. Evaluate the following by the direct application of (62) : 

pb 1)2 _ a 2 pb 

(a) j xdx = , (/3) / e x dx — & — e a . 

J a 2 J a 

Take equal intervals and use the rules for arithmetic and geometric progressions. 

r b 1 r h 1 

7. Evaluate (or) | x m dx = (b m + 1 — a m + 1 ), (j8) I c x dx = (c b — c a ). 

J a m + 1 J a logc 

In the first the intervals should be taken in geometric progression with r n = b/a. 



30 INTRODUCTORY REVIEW 

8. Show directly that (a) f sm 2 xdx = \ ir, (/3) f cos n xdx = 0, if n is odd. 

^o Jo 

9. With the aid of the trigonometric formulas 

cosx + cos2x + • • • + cos(n — l)x = \ [sinwx cot|x — 1 — cosnx], 
sin x + sin 2 x + • • • + sin (n—V)x = \ [(1 — cos nx) cot | x — sin nx], 

J'* b s*b 

cos xdx = sin b — sin a, (/3) J sin xdx = cos a — cos &. 

10. A function is said to be even if /(— x) =f(x) and odd if /(— x) = — /(x) # 

Show (a) f /(a) dx = 2 f 7(x) d^ / even, (/3) f 7(a) dx = 0, / odd. 

11. Show that if an integral is regarded as a function of the lower limit, the 
upper limit being fixed, then 



*» - T- f h ^ x ) dx = ~ ^( a )' if * («) = f A*) 

act J a J a 



(?./:. 



12. Use the relation between definite and indefinite integrals to compare 



L 



f{x)dx = (b-a)f(Z) and F(b)- F(a) = (b- a)F'(£), 



the Theorem of the Mean for derivatives and for definite integrals. 
13. From consideration of Exs. 12 and 4 establish Cauchy's Formula 
AF _ F(b)- F(a) _F'(£) 



a < | < 6, 



A$ $ (6) _ $ (a) <£>'(£) 
which states that the quotient of the increments A.F 1 and A<I> of two functions, in 
any interval in which the derivative <£'(x) does not vanish, is equal to the quotient 
of the derivatives of the functions for some interior point of the interval. ' What 
would the application of the Theorem of the Mean for derivatives to numerator 
and denominator of the left-hand fraction give, and wherein does it differ from 
Cauchy's Formula ? 

14. Discuss the volume of revolution of y =/(x) as the limit of the sum of thin 
cylinders and compare the results with those found in Ex. 9, p. 10. * 

15. Show that the mass of a rod running from a to b along the x-axis is 
I k (b 2 — a 2 ) if the density varies as the distance from the origin (k is a factor of 
proportionality). 

16. Show (a) that the mass in a rod running from a to b is the same as the area 
under the curve y = p(x) between the ordinates x = a and x = 6, and explain why 
this should be seen intuitively to be so. Show (/3) that if the density in a plane slab 
bounded by the x-axis, the curve y =/(x), and the ordinates x = a and x = b is a 

r b 
function p (x) of x alone, the mass of the slab is I yp (x) dx ; also (7) that the mass 

r b 

of the corresponding volume of revolution is / 7ry 2 p (x) dx. 

17. An isosceles triangle has the altitude a and the base 2 5. Find (a) the mass 
on the assumption that the density varies as the distance from the vertex (meas- 
ured along the altitude). Find (/3) the mass of the cone of revolution formed by 
revolving the triangle about its altitude if the law of density is the same. 



FUNDAMENTAL RULES 31 

18. In a plane, the moment of inertia I of a particle of mass m with respect to a 
point is defined as the product mr 2 of the mass by the square of its distance from the 
point. Extend this definition from particles to bodies. 

(a) Show that the moments of inertia of a rod running from a to b and of a 
circular slab of radius a are respectively 

I=( x 2 p (x) dx and I = f 2 7rr s p (r) dr, p the density, 

if the point of reference for the rod is the origin and for the slab is the center. 

(/3) Show that for a rod of length 21 and of uniform density, 1= ^Ml? with 
respect to the center and I = f Ml 2 with respect to the end, M being the total mass 
of the rod. 

(7) For a uniform circular slab with respect to the center I = i Ma 2 . 

(8) For a uniform rod of length 2 1 with respect to a point at a distance d from 
its center is I = M{\ I 2 + d 2 ). Take the rod along the axis and let the point be 
(a, /3) with d 2 = a 2 + /3 2 . 

19. A rectangular gate holds in check the water in a reservoir. If the gate is 
submerged over a vertical distance H and has a breadth B and the top of the 
gate is a units below the surface of the water, find the pressure on the gate. At 
what depth in the water is the point where the pressure is the mean pressure 
over the gate ? 

20. A dam is in the form of an isosceles trapezoid 100 ft. along the top (which 
is at the water level) and 60 ft. along the bottom and 30 ft. high. Find the pres- 
sure in tons. 

21. Find the pressure on a circular gate in a water main if the radius of the 
circle is r and the depth of the center of the circle below the water level is d = r. 

22. In space, moments of inertia are defined relative to an axis and in the for- 
mula I = mr 2 , for a single particle, r is the perpendicular distance from the 
particle to the axis. 

(a) Show that if the density in a solid of revolution generated by y =f(x) varies 
only with the distance along the axis, the moment of inertia about the axis of 

r b 
revolution is I = I h try^p (x) dx. Apply Ex. 18 after dividing the solid into disks. 

J a 

(/3) Find the moment of inertia of a sphere about a diameter in case the density 
is constant ; I — f Ma 2 = T R - 7773a 5 . 

(7) Apply the result to find the moment of inertia of a spherical shell with 
external and internal radii a and b ; I = f If (a 5 — ¥ J )/(a z — 6 3 ). Let b = a and 
thus find I = I Ma? as the moment of inertia of a spherical surface (shell of negli- 
gible thickness). 

(5) For a cone of revolution I — T \ Ma 2 where a is the radius of the base. 

23. If the force of attraction exerted by a mass m upon a point is kmf(r) where 
r is the distance from the mass to the point, show that the attraction exerted at 
the origin by a rod of density p (x) running from a to b along the x-axis is 

A = / kf(x) p (x) dx, and that A = kM/ab, 2£ = p(b— «), 

J a 

is the attraction of a uniform rod if the law is the Law of Nature, that is, 
/(»•) = lA 2 - 



32 INTRODUCTORY REVIEW 

24. Suppose that the density p in the slab of Ex. 16 were a function p (x, y) of 
both x and y. Show that the mass of a small slice over the interval Ax* would be 
of the form 

nV=f(£) nb nbV ny=f(x) "1 

Ax I p (x, y)dy = $ (£) Ax and that I <l> (x) Ax = I I p (x, y) dy dx 

would be the expression for the total mass and would require an integration with 
respect to y in which x was held constant, a substitution of the limits f(x) and 
for y, and then an integration with respect to x from a to b. 

25. Apply the considerations of Ex. 24 to finding moments of inertia of 
(a) a uniform triangle y = rax, y = 0, x = a with respect to the origin, 
(/3) a uniform rectangle with respect to the center, 

(7) a uniform ellipse with respect to the center. 

26. Compare Exs. 24 and 16 to treat the volume under the surface z = p(x, y) 
and over the area bounded by y =/(x), y = 0, x — a, x = b. Find the volume 

(a) under z = xy and over y 2 = 4_px, y = 0, x = 0, x = 6, 

(/3) under 2 = x 2 + ?/ 2 and over x 2 + y 2 — a 2 , y — 0. x = 0, x = Q, 

X 2 ?/ 2 z 2 x 2 v 2 

(7) under \- — ^ — = 1 and over ■ 1 = 1, y = 0, x = 0, x = a. 

a 2 b 2 c 2 . a 2 b 2 

27. Discuss sectorial area \ j r 2 d<p in polar coordinates as the limit of the sum 
of small sectors running out from the pole. 

28. Show that the moment of inertia of a uniform circular sector of angle a 

r 4 d(p in polar coordinates. 
-0 

29. Eind the moment of inertia of a uniform (a) lemniscate r 2 = a 2 cos 2 2 
and (/3) cardioid r = a (1 — cos0) with respect to the pole. Also of (7) the circle 
r = 2 a cos and (5) the rose r = a sin 2 and (e) the rose r = a sin 3 0. 






CHAPTER II 

REVIEW OF FUNDAMENTAL THEORY* 

18. Numbers and limits. The concept and theory of real number, 
integral, rational, and irrational, will not be set forth in detail here. 
Some matters, however, which are necessary to the proper understand- 
ing of rigorous methods in analysis must be mentioned ; and numerous 
points of view which are adopted in the study of irrational number 
will be suggested in the text or exercises. 

It is taken for granted that by his earlier work the reader has become familiar 
with the use of real numbers. In particular it is assumed that he is accustomed 
to represent numbers as a scale, that is, by points on a straight line, and that he 
knows that when a line is given and an origin chosen upon it and a unit of measure 
and a positive direction have been chosen, then to each point of the line corre- 
sponds one and only one real number, and conversely. Owing to this correspond- 
ence, that is, owing to the conception of a scale, it is possible to interchange 
statements about numbers with statements about points and hence to obtain a 
more vivid and graphic or a more abstract and arithmetic phraseology as may be 
desired. Thus instead of saying that the numbers Xi, x 2 , ■ • • are increasing algebra- 
ically, one may say that the points (whose coordinates are) Xi, x 2 , • • • are moving 
in the positive direction or to the right ; with a similar correlation of a decreasing 
suite of numbers with points moving in the negative direction or to the left. It 
should be remembered, however, that whether a statement is couched in geometric 
or algebraic terms, it is always a statement concerning numbers when one has in 
mind the point of view of pure analysis.! 

It may be recalled that arithmetic begins with the integers, including 0, and 
with addition and multiplication. That second, the rational numbers of the . 
form p/q are introduced with the operation of division and the negative rational 
numbers with the operation of subtraction. Finally, the irrational numbers are 
introduced by various processes. Thus V2 occurs in geometry through the 
necessity of expressing the length of the diagonal of a square, and V3 for the 
diagonal of a cube. Again, it is needed for the ratio of circumference to diameter 
in a circle. In algebra any equation of odd degree has at least one real root and 
hence may be regarded as defining a number. But there is an essential difference 
between rational and irrational numbers in that any rational number is of the 

* The object of this chapter is to set forth systematically, with attention to precision 
of statement and accuracy of proof, those fundamental definitions and theorems which 
lie at the basis of calculus and which have been given in the previous chapter from an 
intuitive rather than a critical point of view. 

t Some illustrative graphs will be given ; the student should make many others. 

33 



34 INTRODUCTORY REVIEW 

form ± p/q with q ^ and can therefore be written down explicitly ; whereas 
the irrational numbers arise by a variety of processes and, although they may be 
represented to any desired accuracy by a decimal, they cannot all be written 
down explicitly. It is therefore necessary to have some definite axioms regulating 
the essential properties of irrational numbers. The particular axiom upon which 
stress will here be laid is the axiom of continuity, the use of which is essential 
to the proof of elementary theorems on limits. 

19. Axiom of Continuity. If all the points of a line are divided into 
two classes such that every point of the first class precedes every point of 
the second class, there must be a point C such that any point preceding 
C is in the first class and any pjoint succeeding C is in the second class. 
This principle may be stated in terms of numbers, as : If all real num- 
bers be assorted into two classes such that every number of the first class 
is algebraically less than every number of the second class, there must be 
a member N such that any number less than N is in the first class and 
any number greater than N is in the second. The number N (or point C) 
is called the frontier number (or point), or simply the frontier of the 
two classes, and in particular it is the upjper frontier for the first class 
and the lower frontier for the second. 

To consider a particular case, let all the negative numbers and zero constitute 
the first class and all the positive numbers the second, or let the negative numbers 
alone be the first class and the positive numbers with zero the second. In either 
case it is clear that the classes satisfy the conditions of the axiom and that zero is 
the frontier number such that any lesser number is in the first class and any 
greater in the second. If, however, one were to consider the system of all positive 
and negative numbers but without zero, it is clear that there would be no number 
JSf which would satisfy the conditions demanded by the axiom when the two 
classes were the negative and positive numbers ; for no matter how small a posi- 
tive number were taken as JSf, there would be smaller numbers which would also 
be positive and would not belong to the first class ; and similarly in case it were 
attempted to find a negative JSf. Thus the axiom insures the presence of zero in 
the system, and in like manner insures the presence of every other number — a 
matter which is of importance because there is no way of writing all (irrational) 
numbers in explicit form. 

Further to appreciate the continuity of the number scale, consider the four 
significations attributable to the phrase xx the interval from a to 6." They are 

a = x^=b, a <x = b, a = x < 6, a <x <b. 

That is to say, both end points or either or neither may belong to the interval. In 
the case a is absent, the interval has no first point ; and if b is absent, there is no 
last point. Thus if zero is not counted as a positive number, there is no least 
positive number ; for if any least number were named, half of it would surely be 
less, and hence the absurdity. The axiom of continuity shows that if all numbers 
be divided into two classes as required, there must be either a greatest in the first 
class or a least in the second — the frontier — but not both unless the frontier is 
counted twice, once in each class. 



FUNDAMENTAL THEORY 35 

20. Definition of a Limit. If x is a variable which takes on succes- 
sive values x v x 2 , ■ • -, x { , x } -, • • -, the variable x is said to approach the con- 
stant I as a limit if the numerical difference between x and I ultimately 
becomes, and for all succeeding values of x remains, 
less than any preassigned number no matter how A j x x.- '"'/"' ' x 
small. The numerical difference between x and I 
is denoted by \x — l\ or \l — x\ and is called the absolute value of the 
difference. The fact of the approach to a limit may be stated as 

\x — l\ < e for all cc's subsequent to some x 
or x — I -\- rj, \rj\ < e for all x's subsequent to some x, 

where e is a positive number which may be assigned at pleasure and 
must be assigned before the attempt be made to find an x such that 
for all subsequent x's the relation \x — l\ < e holds. 

So long as the conditions required in the definition of a limit are satisfied there 
is no need of bothering about how the variable approaches its limit, whether from 
one side or alternately from one side and the other, whether discontinuously as in 
the case of the area of the polygons used for computing the area of a circle or 
continuously as in the case of a train brought to rest by its brakes. To speak 
geometrically, a point x which changes its position upon a line approaches the 
point I as a limit if the point x ultimately comes into and remains in an assigned 
interval, no matter how small, surrounding I. 

A variable is said to become infinite if the numerical value of the 
variable ultimately becomes and remains greater than any preassigned 
number K, no matter how large. * The notation is x = ao, but had best 
be read " x becomes infinite," not " x equals infinity." 

Theorem 1. If a variable is always increasing, it either becomes 
infinite or approaches a limit. 



l &? 



That the variable may increase indefinitely is apparent. But if it does not 
become infinite, there must be numbers K which are greater than any value of 
the variable. Then any number must satisfy one of two conditions : either there 
are values of the variable which are greater than it or there are no values of the 
variable greater than it. Moreover all numbers that satisfy the first condition are 
less than any number which satisfies the second. All numbers are therefore 
divided into two classes fulfilling the requirements of the axiom of continuity, and 
there must be a number N such that there, are values of the variable greater than 
any number N — e which is less than N. Hence if e be assigned, there is a value of 
the variable which lies in the interval N — e < x ==i iV, and as the variable is always 
increasing, all subsequent values must lie in this interval. Therefore the variable 
approaches JSf as a limit. 

* This definition means what it says, and no more. Later, additional or different 
meanings may be assigned to infinity, but not now. Loose and extraneous concepts in 

this connection are almost certain to introduce errors and confusion. 



36 INTRODUCTORY REVIEW 

EXERCISES 

1. If Xi, x 2 , • • •, x n , • • •, x n + p , • • • is a suite approaching a limit, apply the defi- 
nition of a limit to show that when e is given "it must be possible to find a value of 
n so great that \x n+p — x n \ < e for all values of p. 

2. If #i, x 2 , • • • is a suite approaching a limit and if y 1? y 2 , - ■ • is any suite such 
that |y w — x„ ] approaches zero when n becomes infinite, show that the ?/'s approach 
a limit which is identical with the limit of the x's. 

3. As the definition of a limit is phrased in terms of inequalities and absolute 
values, note the following rules of operation : 

(a) If a > and c > 6, then - > - and - < - , 

a a c b 

(£) \a+.b + c + :...|s|d'|,+ |6| + |c]-+ •••, (7) [a&c...| = |a|.[5].[c|-.., 

where the equality sign in (/3) holds only if the numbers a, b, c, • • • have the same 
sign. By these relations and the definition of a limit prove the fundamental 
theorems : 

If lim x = X and lim y = F, then lim (x ± y) = X ± Y and lim xy = XT. 

4. Prove Theorem 1 when restated in the slightly changed form : If a variable 
x never decreases and never exceeds K, then x approaches a limit N and X = K. 
Illustrate fully. State and prove the corresponding theorem for the case of a 
variable never increasing. 

5. If Xi, x 2 , • • • and ?/ l7 y 2 , ■ • ■ are two suites of which the first never decreases 
and the second never increases, all the y's being greater than any of the x's, and if 
when e is assigned an n can be found such that y n — x n < e, show that the limits 
of the suites are identical. 

6. If Xi, x 2 , • • • and y h y 2 , ■ • • are two suites which never decrease, show by Ex. 4 
(not by Ex. 3) that the suites X\ + y h x 2 + y 2 , • • • and x x ?/i, x 2 y 2 , ■ ■ • approach 
limits. Note that two infinite decimals are precisely two suites which never de- 
crease as more and more figures are taken. They do not always increase, for some 
of the figures may be 0. 

7. If the word ff all " in the hypothesis of the axiom of continuity be assumed to 
refer only to rational numbers so that the statement becomes : If all rational 
numbers be divided into two classes • • • , there shall be a number N (not neces- 
sarily rational) such that • • • ; then the conclusion may be taken as defining a 
number as the frontier of a sequence of rational numbers. Show that if two num- 
bers JT, Y be defined by two such sequences, and if the sum of the numbers be 
defined as the number defined by the sequence of the sums of corresponding terms 
as in Ex. 6, and if the product of the numbers be defined as the number defined by 
the sequence of the products as in Ex. 6, then the fundamental rules 

X +Y=Y+ X, XY = YX, (X +Y)Z = XZ + YZ 

of arithmetic hold for the numbers X, Y, Z defined by sequences. In this way a 
complete theory of irrationals may be built up from the properties of rationals 
combined with the principle of continuity, namely, 1° by defining irrationals as 
frontiers of sequences of rationals, 2° by defining the operations of addition, multi- 
plication, • • • as operations upon the rational numbers in the sequences, 3° by 
showing that the fundamental rules of arithmetic still hold for the irrationals. 






FUNDAMENTAL THEOEY 37 

8. Apply the principle of continuity to show that there is a positive number x 
such that x 2 = 2. To do this it should be shown that the rationals are divisible 
into two classes, those whose square is less than 2 and those whose square is not 
less than 2 ; and that these classes satisfy the requirements of the axiom of conti- 
nuity. In like manner if a is any positive number and n is any positive integer, 
show that there is an x such that x n = a. 

21. Theorems on limits and on sets of points. The theorem on 
limits which is of fundamental algebraic importance is 

Theorem 2. If R (x, y, z, ■ • •) be any rational function of the variables 
x, y, z, ■ ■ ■ . and if these variables are approaching limits X, Y, Z, • • • , 
then the value of R approaches a limit and the limit is R (X, Y, Z, • • ■), 
provided there is no division by zero. 

As any rational expression is made up from its elements by combinations of 
addition, subtraction, multiplication, and division, it is sufficient to prove the 
theorem for these four operations. All except the last have been indicated in the 
above Ex. 3. As multiplication has been cared for, division need be considered 
only in the simple case of a reciprocal 1/x. It must be proved that if lim x = A", 
then lim (1/x) = 1/X. Xow 

1 __1_ 
x~~X 



— ', by Ex. 3 (7) above. 



This quantity must be shown to be less than any assigned e. As the quantity is 

complicated it will be replaced by a simpler one which is greater, owing to an 

increase in the denominator. Since x = X, x — X may be made numerically as 

small as desired, say less than e', for all x's subsequent to some particular x. Hence 

if e' be taken at least as small as ||X|, it appears that |x| must be greater than 

l-\X\. Then 

I X — X I I X — X I e . ^ _ . . , 

' ! < ! ' = , bv Ex. 3 la) above, 

\x\\X\ i[A|2 i|AT ~ K J 

and if e' be restricted to being less than ||A| 2 e, the difference is less than e and 
the theorem that lim (1/x) = 1/X is proved, and also Theorem 2. The necessity 
for the restriction X jt and the corresponding restriction in the statement of 
the theorem is obvious. 

Theorem 3. If when e is given, no matter how small, it is possible 
to find a value of n so great that the difference \x n+p — x n \ between x n 
and every subsequent term x n+p in the suite x 1} x 2 , ■■■, x n , ■ • • is less 
than e. the suite approaches a limit, and conversely. 

The converse part has already been given as Ex. 1 above. The theorem itself is 
a consequence of the axiom of continuity. First note that as \x n + p — x n \ < e for 
all x's subsequent to x n . the x's cannot become infinite. Suppose 1° that there 
is some number I such that no matter how remote x n is in the suite, there are 
always subsequent values of x which are greater than / and others which are less 
than I. As all the x's after x n lie in the interval 2 e and as I is less than some x's 
and greater than others, I must lie in that interval. Hence \l — x n+p \ < 2 e for all 



38 INTRODUCTORY REVIEW 

x's subsequent to x n . But now 2 e can be made as small as desired because e can be 
taken as small as" desired. Hence the definition of a limit applies and the x's 
approach I as a limit. 

Suppose 2° that there is no such number I. Then every number k is such that 
either it is possible to go so far in the suite that all subsequent numbers x are 
as great as k or it is possible to go so far that all subsequent x's, are less than k. 
Hence all numbers k are divided into two classes which satisfy the requirements of 
the axiom of continuity, and there must be a number N such that the x's ultimately 
come to lie between JSf — e / and N + e', no matter how small e is. Hence the sc's 
approach N as a limit. Thus under either supposition the suite approaches a limit 
and the theorem is proved. It may be noted that under the second supposition the 
x's ultimately lie entirely upon one side of the point N and that the condition 
\x n + p — x n \-< e is not used except to show that the x's remain finite. 

22. Consider next a set of points (or their correlative numbers) 
without any implication that they form a suite, that is, that one may 
be said to be subsequent to another. If there is only a finite number 
of points in the set, there is a point farthest to the right and one 
farthest to the left. If there is an infinity of points in the set, two 
possibilities arise. Either 1° it is not possible to assign a point K so 
far to the right that no point of the set is farther to the right — in 
which case the set is said to be unlimited above — or 2° there is a 
point K such that no point of the set is beyond K — and the set is 
said to be limited above. Similarly, a set may be limited below or un- 
limited below. If a set is limited above and below so that it is entirely 
contained in a finite interval, it is said merely to be limited. If there 
is a point C such that in any interval, no matter how small, surround- 
ing C there are points of the set, then C is called a point of condensa- 
tion of the set (C itself may or may not belong to the set). 

Theorem 4. Any infinite set of points which is limited has an 
upper frontier (maximum ?), a lower frontier (minimum ?), and at 
least one point of condensation. 

Before proving this theorem, consider three infinite sets as illustrations : 
(a) 1, 1.9, 1.99, 1.999, • • -, (/3) - 2, • • ., - 1.99, - 1.9, - 1, 

(7) — 1, — h ~ b ' * * > h h !• 

In (a) the element 1 is the minimum and serves also as the lower frontier ; it is 
clearly not a point of condensation, but is isolated. There is no maximum ; but 2 
is the upper frontier and also a point of condensation. In (/3) there is a maximum 
— 1 and a minimum — 2 (for — 2 has been incorporated with the set). In (7) there 
is a maximum and minimum ; the point of condensation is 0. If one could be sure 
that an infinite set had a maximum and minimum, as is the case with finite 
sets, there would be no need of considering upper and lower frontiers. It is clear 
that if the upper or lower frontier belongs to the set, there is a maximum or 
minimum and the frontier is not necessarily a point of condensation ; whereas 



FUNDAMENTAL THEORY 39 

if the frontier does not belong to the set, it is necessarily a point of condensation and 
the corresponding extreme point is missing. 

To prove that there is an upper frontier, divide the points of the line into two 
classes, one consisting of points which are to the left of some point of the set, the 
other of points which are not to the left of any point of the set — then apply the 
axiom. Similarly for the lower frontier. To show the existence of a point of con- 
densation, note that as there is an infinity of elements in the set, any point p is such 
that either there is an infinity of points of the set to the right of it or there is not. 
Hence the two classes into which all points are to be assorted are suggested, and 
the application of the axiom offers no difficulty. 

EXERCISES 

1. In a manner analogous to the proof of Theorem 2, show that 

{a ) Km ZZLL = h, (3) lim ^Lll = 5, ( 7 ) li m X l±l = _ lm 

2. Given an infinite series S = U\ -f u 2 + u s + • • • . Construct the suite 

S 1 = Mi, S 2 = Ul + U 2 , £3 = Ui + U 2 + U S , • • • , S{ = tti + U 2 + • • • + Mi, • • • , 

where Si is the sum of the first i terms. Show that Theorem 3 gives : The neces- 
sary and sufficient condition that the series S converge is that it is possible to find 
an n so large that \S n+p — S n \ shall be less than an assigned e for all values of p. 
It is to be understood that a series converges when the suite of S's approaches a limit,. 
and conversely. 

3. If in a series Ui — u 2 + u s — u± + ■ ■ • the terms approach the limit 0, are 
alternately positive and negative, and each term is less than the preceding, the 
series converges. Consider the suites Si, S s , S 5 , • • • and S 2 , *S 4 , Sq, • • • . 

4. Given three infinite suites of numbers 

aJi, «2, • • •, En, • • • ; yi, y 2 , • • •, y n , • • • ; zi, z 2 , ■ ■ •, z n , ■ ■ ■ 
of which the first never decreases, the second never increases, and the terms of the 
third lie between corresponding terms of the first two, x n == z n ^ y n . Show that 
the suite of z's has a point of condensation at or between the limits approached by 
the x's and by the y's ; and that if lim x — lim y = I, then the z's approach I as a 
limit. 

5. Restate the definitions and theorems on sets of points in arithmetic terms. 

6. Give the details of the proof of Theorem 4. Show that the proof as outlined 
gives the least point of condensation. How would the proof be worded so as to give 
the greatest point of condensation ? Show that if a set is limited above, it has an 
upper frontier but need not have a lower frontier. 

7. If a set of points is such that between any two there is a third, the set is said 
to be dense. Show that the rationals form a dense set ; also the irrationals. Show 
that any point of a dense set is a point of condensation for the set. 

8. Show that the rationals p/q where q < K do not form a dense set — in fact 
are a finite set in any limited interval. Hence in regarding any irrational as the 
limit of a set of rationals it is necessary that the denominators and also the numer- 
ators should become infinite. 



40 INTRODUCTORY REVIEW 

9. Show that if an infinite set of points lies in a limited region of the plane, 
say in the rectangle a ==i x ==i 6, c =§ y = d, there must be at least one point of 
condensation of the set. Give the necessary definitions and apply the axiom 
of continuity successively to the abscissas and orclinates. 

23. Real functions of a real variable. If x be a variable which 
takes on a certain set of values of which the totality may be denoted 
by \_x~\ and if y is a second variable the value of which is uniquely 
determined for each x of the set [cc], then y is said to be a function of 
x defined over the set [x~\. The terms " limited," " unlimited," " limited 
above," " unlimited below," • • • are applied to a function if they are 
applicable to the set [_y~\ of values of the function. Hence Theorem 4 
has the corollary : 

Theorem 5. If a function is limited over the set [cc], it has an 
upper frontier M and a lower frontier m for that set. 

If the function takes on its upper frontier ill, that is, if there is a 
value x Q in the set [x~\ such that f(x Q ) = M, the function has the abso- 
lute maximum M at x Q ; and similarly with respect to the lower 
frontier. In any case, the difference M — m between the upper and 
lower frontiers is called the oscillation of the function for the set [a;]. 
The set [jc] is generally an interval. 

Consider some illustrations of functions and sets over which they are defined. 
The reciprocal 1/x is defined for all values of x save 0. In the neighborhood of 
the function is unlimited above for positive x\s and unlimited below for negative x's. 
It should be noted that the function is not limited in the interval < x =§= a but is 
limited in the interval e == x ^ a where e is any assigned positive number. The 
function + vx is defined for all positive x's including and is limited below. It 
is not limited above for the totality of all positive numbers ; but if K is assigned, 
the function is limited in the interval ^ x ^ K . The factorial function x ! is de- 
fined only for positive integers, is limited below by the value 1, but is not limited 
above unless the set [x] is limited above. The function E (x) denoting the integer 
not greater than x or "the integral part of x " is defined for all positive numbers 
— for instance E (3) — E{tt) = 3. This function is not expressed, like the elemen- 
tary functions of calculus, as a "formula " ; it is defined by a definite law, however, 
and is just as much of a function as x 2 + 3x -f 2 or \ sin 2 2x + logx. Indeed it 
should be noted that the elementary functions themselves are in the first instance 
defined by definite laws and that it is not until after they have been made the 
subject of considerable study and have been largely developed along analytic lines 
that they appear as formulas. The ideas of function and formula are essentially 
distinct and the latter is essentially secondary to the former. 

The definition of function as given above excludes the so-called multiple-valued 
functions such as v x and sin -1 x where to a given value of x correspond more than 
one value of the function. It is usual, however, in treating multiple-valued func- 
tions to resolve the functions into different parts or branches so that each branch 
is a single-valued function. Thus +vx is one branch and — Vx the other branch 



FUNDAMENTAL THEORY 41 

of Vx ; in fact when x is positive the symbol Vx is usually restricted to mean 
merely + Vx and thus becomes a single-valued symbol. One branch of sin -1 x con- 
sists of the values between — \ it and + \ 7r, other branches give values between 
\ it and f 7r or — \tc and — f 7r, and so on. Hence the term "function" will be 
restricted in this chapter to the single-valued functions allowed by the definition. 

24. If x = a is any point of an interval over which f(x) is defined, 
the function f(x) is said to be continuous at the point x = a if 

lim/(cc) =/(«), no matter how x — a. 

x = a 

The function is said to be continuous in the interval if it is continuous 
at every point of the interval. If the function is not continuous at the 
point a, it is said to be discontinuous at a ; and if it fails to be con- 
tinuous at any one point of an interval, it is said to be discontinuous 
in the interval. 

Theorem 6. If any finite number of functions are continuous (at a 
point or over an interval), any rational expression formed of those 
functions is continuous (at the point or over the interval) provided no 
division by zero is called for. 

Theorem 7. If y=f(x) is continuous at x Q and takes the value 
y =f(x Q ) and if z = <f>(y) is a continuous function of y at y — y , then 
z = <j>\_f(x)~] will be a continuous function of x at x Q . 

In regard to the definition of continuity note that a function cannot be con- 
tinuous at a point unless it is defined at that point. Thus e -1 /* 2 is not continuous 
at x = because division by is impossible and the function is undefined. If, how- 
ever, the function be defined at as/(0) =0, the function becomes continuous at 
x = 0. In like manner the function 1/x is not continuous at the origin, and in this 
case it is impossible to assign to/(0) any value which will render the function 
continuous ; the. function becomes infinite at the origin and the very idea of be- 
coming infinite precludes the possibility of approach to a definite limit. Again, the 
function E (x) is in general continuous, but is discontinuous for integral values 
of x. When a function is discontinuous at x = a, the amount of the discontinuity is 
the limit of the oscillation M — m of the function in the interval a — 5 <x < a + 5 
surrounding the point a when 5 approaches zero as its limit. The discontinuity 
of E (x) at each integral value of x is clearly 1 ; that of 1/x at the origin is infi- 
nite no matter what value is assigned to /(0). 

In case the interval over which /(x) is defined has end points, say a = x == 6, 
the question of continuity at x = a must of course be decided by allowing x to 
approach a from the right-hand side only ; and similarly it is a question of left- 
handed approach to b. In general, if for any reason it is desired to restrict the 
approach of a variable to its limit to being one-sided, the notations x = a+ and 
x = b~ respectively are used to denote approach through greater values (right- 
handed) and through lesser values (left-handed). It is not necessary to make this 
specification in the case of the ends of an interval ; for it is understood that x 
shall take on only values in the interval in question. It should be noted that 



42 INTRODUCTORY REVIEW 

lim f(x) =f(x ) when x = x + in no wise implies the continuity of f(x) at x ; a 
simple example is that of E (x) at the positive integral points. 

The proof of Theorem 6 is an immediate corollary application of Theorem 2. For 

lim R [/(x), (x) .-.'•]=. 22 [lim/(x), lim (x), • • •] = R [/(lim x), (lim x), • • •], 

and the proof of Theorem 7 is equally simple. 

Theorem 8. If f(x) is continuous at x = «, then for any positive 
e which has been assigned, no matter how small, there may be found a 
number 8 such that \f(x)—f(d)\<e in the interval \x— a\<8, and 
hence in this interval the oscillation of f(x) is less than 2 e. And 
conversely, if these conditions hold, the function is continuous. 

This theorem is in reality nothing hut a restatement of the definition of conti- 
nuity combined with the definition of a limit. For "lim/(x) =f(a) when x = a, 
no matter how" means that the difference between /(x) and /(a) can be made as 
small as desired by taking x sufficiently near to a ; and conversely. The reason 
for this restatement is that the present form is more amenable to analytic opera- 
tions. It also suggests the geometric picture which corre- 
sponds to the usual idea of continuity in graphs. Tor the 
theorem states that if the two lines y =f{a) ± e be drawn, 
the graph of the function remains between them for at least 
the short distance 5 on each side of x = a ; and as e may be 
assigned a value as small as desired, the graph cannot exhibit 
breaks. On the other hand it should be noted that the actual 
physical graph is not a curve but a band, a two-dimensional region of greater or 
less breadth, and that a function could be discontinuous at every point of an 
interval and yet lie entirely within the limits of any given physical graph. 

It is clear that 5, which has to be determined subsequently to e, is in general 
more and more restricted as e is taken smaller and that for different points it is 
more restricted as the graph rises more rapidly. Thus if f(x) = 1/x and e = 1/1000, 
5 can be nearly 1/10 if x = 100, but must be slightly less than 1/1000 if x = 1, and 
something less than 10- 6 if x is 10- 3 . Indeed, if x be allowed to approach zero, the 
value 5 for any assigned e also approaches zero ; and although the function 
f(x) = 1/x is continuous in the interval < x === 1 and for any given x and e a 
number 5 may be found such that |/(x) — /(x ) | < e when |x — x 1 < 5, yet it is not 
possible to assign a number 8 which shall serve uniformly for all values of x . 

25. Theorem 9. If a function f(x) is continuous in an interval 
a p= x ^ b with end points, it is possible to find a 8 such that 
\f(x) —f(x )\<e when \x — x 1 < 8 for all points x ; and the function 
is said to be uniformly continuous. 

The proof is conducted by the method of reductio ad absurdum. Suppose e 
is assigned. Consider the suite of values i i, i, • • • , or any other suite which 
approaches zero as a limit. Suppose that no one of these values will serve as a 5 
for all points of the interval. Then there must be at least one point for which 1 
will not serve, at least one for which i will not serve, at least one for which \ will 
not serve, and so on indefinitely. This infinite set of points must have at least one 




FUNDAMENTAL THEOKY 43 

point of condensation C such that in any interval surrounding C there are points for 
which 2 - * will not serve as 5, no matter how large Jc. But now by hypothesis /(x) 
is continuous at C and hence a number 5 can be found such that |/(x) — f(C)\< \ e 
when \x — x 1 < 2 8: The oscillation of f(x) in the whole interval 4 5 is less than e. 
Now if x be any point in the middle half of this interval, | x — C \ < 8 ; and if x 
satisfies the relation | x — &o | < 5, it must still lie in the interval 4 8 and the differ- 
ence \f(x) —f(x ) | < e, being surely not greater than the oscillation of /in the whole 
interval. Hence it is possible to surround C with an interval so small that the 
same 5 will serve for any point of the interval. This contradicts the former con- 
clusion, and hence the hypothesis upon which that conclusion was based must have 
been false and it must have been possible to find a 8 which would serve for all 
points of the interval. The reason why the proof would not apply to a function 
like 1/x defined in the interval < x = 1 lacking an. end point is precisely that 
the point of condensation C would be 0, and at the function is not continuous 
and \f(x) —f(C) \<%e, \x — C'|<25 could not be satisfied. 

Theorem 10 r If a function is continuous in a region which includes 
its end points, the function is limited. 

Theorem 11. If a function is continuous in an interval which includes 
its end points, the function takes on its upper frontier and has a maxi- 
mum 31; similarly it has a minimum m. 

These are successive corollaries of Theorem 9. For let e be assigned and let 5 
be determined so as to serve uniformly for all points of the interval. Divide the 
interval b — a into n successive intervals of length 8 or less. Then in each such 
interval / cannot increase by more than e nor decrease by more than e. Hence / 
will be contained between the values /(a) + ne and /(a) — ne, and is limited. And 
f(x) has an upper and a lower frontier in the interval. Next consider the rational 
function l/(M — f) of /. By Theorem 6 this is continuous in the interval unless 
the denominator vanishes, and if continuous it is limited. This, however, is impos- 
sible for the reason that, as M is a frontier of values of /, the difference 31 — / 
may be made as small as desired. Hence 1/(31— f) is not continuous and there 
must be some value of x for which / = M . 

Theorem 12. If f(x) is continuous in the interval a ^= x ^ b with end 
points and if f(a) and f(b) have opposite signs, there is at least one 
point |, a < £ < b, in the interval for which the function vanishes. 
And whether /(«) and f(b) have opposite signs or not, there is a point 
£, a < £ < b, such that /(£) = /a, where /x is any value intermediate be- 
tween the maximum and minimum of / in the interval. 

For convenience suppose that /(a) < 0. Then in the neighborhood of x — a the 
function will remain negative on account of. its continuity; and in the neighbor- 
hood of b it will remain positive. Let £ be the lower frontier of values of x which 
make/(x) positive. Suppose that/(£) were either positive or negative. Then as 
/ is continuous, an interval could be chosen surrounding £ and so small that / re- 
mained positive or negative in that interval. In neither case could £ be the lower 
frontier of positive values. Hence the contradiction, and /(£) must be zero. To 



44 INTRODUCTORY REVIEW 

prove the second part of the theorem, let c and d be the values of x which make 
/ a minimum and maximum. Then the function f — /x has opposite signs at c and 
d, and must vanish at some point of the interval between c and d ; and hence a 
fortiori at some point of the interval from a to b. 

EXERCISES 

1. Note that x is a continuous function of x, and that consequently it follows 
from Theorem 6 that any rational fraction P(x)/Q(x), where P and Q are poly- 
nomials in x, must be continuous for all x 1 s except roots of Q(x) = 0. 

2. Graph the function x — E (x) for x === and show that it is continuous except 
for integral values of x. Show that it is limited, has a minimum 0, an upper fron- 
tier 1, but no maximum. 

3. Suppose that/(x) is defined for an infinite set [x] of which x = a is a point 
of condensation (not necessarily itself a point of the set). Suppose 

lim [f(x')-f(x")] = or |/(x') _/(x")|<e, \x' - a\< .5, \x" - a\< 5, 

a/, x" = a 

when x' and x" regarded as independent variables approach a as a limit (passing 
only over values of the set [x], of pourse). Show that/(x) approaches a limit as 
x = a. By considering the set of values of /(x), the method of Theorem 3 applies 
almost verbatim. Show that there is no essential change in the proof if it be 
assumed that x" and x" become infinite, the set [x] being unlimited instead of 
having a point of condensation a. 

4. From the formula sin x < x and the formulas for sin u — sin v and cos u — cos v 
show that A sin x and A cos x are numerically less than 2 1 Ax | ; hence infer that sin x 
and cosx are continuous functions of x for all values of x. 

5. What are the intervals of continuity for tanx and cscx ? If e = 10- 4 , what 
are approximately the largest available values of 5 that will make |/(x) — /(x ) | <e 
when x = 1°, 30°, 60°, 89° for each ? Use a four-place table. 

6. Let f(x) be defined in the interval from to 1 as equal to when x is irra- 
tional and equal to 1/q when x is rational and expressed as a fraction p/q in lowest 
terms. Show that/ is continuous for irrational values and discontinuous for 
rational values. Ex. 8, p. 39, will be of assistance in treating the irrational values. 

7. Note that in the definition of continuity a generalization may be introduced 
by allowing the set [x] over which / is defined to be any set each point of which 
is a point of condensation of the set, and that hence continuity over a dense set 
(Ex. 7 above), say the rationals or irrationals, may be defined. This is important 
because many functions are in the first instance defined only for rationals and are 
subsequently defined for irrationals by interpolation. Note that if a function is 
continuous over a dense set (say, the rationals), it does not follow that it is uni- 
formly continuous over the set. Eor the point of condensation C which was used 
in the proof of Theorem 9 may not be a point of the set (may be irrational), and 
the proof would fall through for the same reason that it would in the case of 1/x 
in the interval < x === 1, namely, because it could not be affirmed that the function 
was continuous at C. Show that if a function is defined and is uniformly continu- 
ous over a dense set, the value /(x) will approach a limit when x approaches any 
value a (not necessarily of the set, but situated between the upper and lower 



FUNDAMENTAL THEORY 45 

frontiers of the set), and that if this limit be defined as the value of /(a), the 
function will remain continuous. Ex. 3 may be used to advantage. 

8. By factoring (x + Ax) n — x n , show for integral values of n that when 
=== x === K , then A (x n ) < nK n - 1 Ax for small Ax's and consequently x 11 is uniformly 
continuous in the interval ==E x S§ K . If it be assumed that x n has been defined 
only for rational x's, it follows from Ex. 7 that the definition may be extended 
to all x's and that the resulting x n will be continuous. 

9. Suppose (a) that/(x) +f{y) =f(x + y) for any numbers x and y. Show that 
f(n) = nf(l) and nf(l/n) =/(l), and hence infer that f(x) = xf(l) = Cx, where 
C=/(l), for all rational x's. Erom Ex. 7 it follows that if f(x) is continuous, 
f(x) = Cx for all x's. Consider (/3) the function /(x) such that/(x) f(y) =f(x + y). 
Show that it is Ce x = a x . 

10. Show by Theorem 12 that if y =/(x) is a continuous constantly increasing 
function in the interval a =e= x === 6, then to each value of y corresponds a single value 
of x so that the function x =/ _1 (y) exists and is single-valued ; show also that 
it is continuous and constantly increasing. State the corresponding theorem if 
f(x) is constantly decreasing. The function f~ 1 (y) is called the inverse function 
to/(x). 

11. Apply Ex. 10 to discuss y = Vx, where n is integral, x is positive, and only 
positive roots are taken into consideration. 

12. In arithmetic it may readily be shown that the equations 

a m a n = a m + n , (a m ) n = a mn , a n b n — (ab) n , 

are true when a and b are rational and positive and when m and n are any positive 
and negative integers or zero, (a) Can it be inferred that they hold when a 
and b are positive irrationals ? (/S) How about the extension of the fundamental 
inequalities 

x n > 1, when x > 1, x n < 1, when =§ x < 1 

to all rational values of n and the proof of the inequalities 

x m >x" if m>n and x>l, x m <x n if m>n and 0<x<l. 

(7) Next consider x as held constant and the exponent n as variable. Discuss the 
exponential function a x from this relation, and Exs. 10, 11, and other theorems that 
may seem necessary. Treat the logarithm as the inverse of the exponential. 

26. The derivative. If x = a is a point of an interval over which 
f(x) is defined and if the quotient 

\f f(« + h) -f(h) 

^ = h "' h = Ax > 

approaches a limit when h approaches zero, no matter 7wiv, the function 
f(x) is said to be differ entiable at x = a and the value of the limit of 
the quotient is the derivative f 1 (a) of f at x = a. In the case of differ- 
entiability, the definition of a limit gives 

f(„: + J t )-f(„) =fl(a) + r! o] _ f{a + h) _ f{a) = hfl{a) + rih! (1) 
where lim -q = when lim h = 0, no matter how. 



46 INTEODUCTORY REVIEW 

In other words if e is given, a 5 can be found so that |i;]<e when \h\<5. This 
shows that a function differentiate at a as in (1) is continuous at a. For 

\f(a + h) -f(a)\^ \f'(a)\8+ eS, \h\ < 8. 

If the limit of the quotient exists when h = through positive values only, the 
function has a right-hand derivative which may be denoted by /' (a+) and similarly 
for the left-hand derivative f (a~). At the end points of an interval the derivative 
is always considered as one-handed ; but for interior points the right-hand and left- 
hand derivatives must be equal if the function is to have a derivative (unqualified). 
The function is said to have an infinite derivative at a if the quotient becomes infi- 
nite as h = ; but if a is an interior point, the quotient must become positively 
infinite or negatively infinite for all manners of approach and not positively infinite 
for some and negatively infinite for others. Geometrically this allows a vertical 
tangent with an inflection point, but not with a cusp as in Fig. 3, p. 8. If infinite 
derivatives are allowed, the function may have a derivative and yet be discontin- 
uous, as is suggested by any figure where /(a) is any value between lim/(x) when 
x = a+ and ]\mf(x) when x = a~. 

Theorem 13. If a function takes on its maximum (or minimum) at 
an interior point of the interval of definition and if it is differentiable 
at that point, the derivative is zero. 

Theorem 14. Rollers Theorem. If a function f(x) is continuous over 
an interval a ^ x ^ b with end points and vanishes at the ends and has 
a derivative at each interior point a < x < b, there is some point £, 
a <£ <b, such that/'© = 0. 

Theorem 15. Theorem of the Mean. If a function is continuous over 
an interval a ^ x ^ b and has a derivative at each interior point, there 
is some point £ such that 

b — a li 

where h ^ b — a* and 6 is a proper fraction, < < 1. 

To prove the first theorem, note that if /(a) = M, the difference f(a + h) —f(a) 
cannot be negative for any value of h and the quotient Af/h cannot be positive 
when h > and cannot be negative when h < 0. Hence the right-hand derivative 
cannot be positive and the left-hand derivative cannot be negative. As these two 
must be equal if the function has a derivative, it follows that they must be zero, 
and the derivative is zero. The second theorem is an immediate corollary. For as 
the function is continuous it must have a maximum and a minimum (Theorem 11) 
both of which cannot be zero unless the function is always zero in the interval. 
Now if the function is identically zero, the derivative is identically zero and the 
theorem is true ; whereas if the function is not identically zero, either the maximum 
or minimum must be at an interior point, and at that point the derivative will vanish. 

* That the theorem is true for any part of the interval from a to b if it is true for the 
whole interval follows from the fact that the conditions, namely, that / be continuous 
and that/' exist, hold for any part of the interval if they hold for the whole. 



FUNDAMENTAL THEORY 17 

To prove the last theorem construct the auxiliary function 

iw =/(x)-/ (a) - (x- a) ^M, ? ( x) =/W -fflzM. 

& — a 6 — a 

As ^ (a) = ^ (&) = 0, Rolle's Theorem shows that there is some point for which 
■y (£) = 0, and if this value be substituted in the expression for xp' (x) the solution 
for/'(£) gives the result demanded by the theorem. The proof, however, requires 
the use of the function \]/ (x) and its derivative and is not complete until it is shown 
that \f/(x) really satisfies the conditions of Rolle's Theorem, namely, is continuous 
in the interval a ==§ x =s b and has a derivative for every point a < x < b. The con- 
tinuity is a consequence of Theorem 6 ; that the derivative exists follows from the 
direct application of the definition combined with the assumption that the deriva- 
tive of /exists. 

27. Theorem 16. If a function has a derivative which is identically 
zero in the interval a ^ x ^ b, the function is constant ; and if two 
functions have derivatives equal throughout the interval, the functions 
differ by a constant. 

Theorem 17. If f(x) is differentiable and becomes infinite when 
x = a, the derivative cannot remain finite as x = a. 

Theorem 18. If the derivative f'(x) of a function exists and is a 
continuous function of x in the interval a ^ x ^ b, the quotient Af/h 
converges uniformly toward its limit f (x). 

These theorems are consequences of the Theorem of the Mean. For the first, 

f(a + h)-f(a) = hf(a + 0h) = O, if h^b-a, or f(a+h)=f(a). 

Hence /(x) is constant. And in case of two functions/ and <f> with equal derivatives, 
the difference \p (x) =f(x) — <p (x) will have a derivative that is zero and the differ- 
ence will be constant. Tor the second, let x be a fixed value near a and suppose that 
in the interval from x to a the derivative remained finite, say less than K. Then 

| /(xo + h) - f(x ) ! = [ hf\xo + Oh) \^\h\K. 

Now let x + h approach a and note that the left-hand term becomes infinite and 
the supposition that/' remained finite is contradicted. For the third, note that/', 
being continuous, must be uniformly continuous (Theorem 9), and hence that if e is 
given, a 5 may be found such that 

li 

when \h\< 8 and for all x\s in the interval ; and the theorem is proved. 

Concerning derivatives of higher order no special remarks are necessary. Each 
is the derivative of a definite function — the previous derivative. If the deriva- 
tives of the first n orders exist and are continuous, the derivative of order n + 1 
may or may not exist. In practical applications, however, the functions are gen- 
erally indefinitely differentiable except at certain isolated points. The proof of 
Leibniz's Theorem (§ 8) may be revised so as to depend on elementary processes. 
Let the formula be assumed for a given value of n. The only terms which can 



48 INTRODUCTORY REVIEW 

contribute to the term D i uD n + 1 ~ i v in the formula for the (n + l)st derivative of 
uv are the terms 

n(n — 1) • • • (ft — i + 2) ^ . , ^ ... . n(n — 1) • • • (n — i + 1) _. -^ 

1 • 2 • • • (i - 1) 1 • 2 • • • i 

in which the first factor is to be differentiated in the first and the second in the 
second. The sum of the coefficients obtained by differentiating is 

?i (ft - 1) • • • (ft - i + 2) n (ft - 1) ■ • • (ft - i + 1) _ (ft + 1) n ■ • • (n - i + 2) 
1' 2 • - • (i — 1) 1 • 2 • • • i 1 • 2 • • • i 

which is precisely the proper coefficient for the term _D l 'iii> + 1 ~ i v in the expansion 
of the (ft + 1) st derivative of uv by Leibniz's Theorem. 

With regard to this rule and the other elementary rules of operation (4)-(7) of 
the previous chapter it should be remarked that a theorem as well as a rule is in- 
volved — thus : If two functions u and v are cliff erentiable at x , then the product 
uv is cliff erentiable at x , and the value of the derivative is u (x ) v" (x ) + u' (x ) v (x ) . 
And similar theorems arise in connection with the other rules. As a matter of fact 
the ordinary proof needs only to be gone over with care in order to convert it into 
a rigorous demonstration. But care does need to be exercised both in stating the 
theorem and in looking to the proof. Tor instance, the above theorem concerning 
a product is not true if infinite derivatives are allowed. For let u be — 1, 0, or + 1 
according as x is negative, 0, or positive, and let v = x. Now v has always a deriva- 
tive which is 1 and u has always a derivative which is 0, + co, or according as x 
is negative, 0, or positive. The product uv is |x|, of which the derivative is — 1 for 
negative x 1 s, +1 for positive x's, and nonexistent for 0. Here the product has no 
derivative at 0, although each factor has a derivative, and it would be useless to have 
a formula for attempting to evaluate something that did not exist. 



EXERCISES 

1 . Show that if at a point the derivative of a function exists and is positive, the 
function must be increasing at that point. 

2. Suppose that the derivatives f'{a) and /'(&) exist and are not zero. Show 
that /(a) and f(b) are relative maxima or minima of / in the interval a = x = 6, and 
determine the precise criteria in terms of the signs of the derivatives f'(a) and f'(b). 

3. Show that if a continuous function has a positive right-hand derivative at 
every point of the interval a ==§ x =i &, then f(b) is the maximum value of /. Simi- 
larly, if the right-hand derivative is negative, show that/(6) is the minimum of/. 

4. Apply the Theorem of the Mean to show that if f'{x) is continuous at a, then 

lim /(*0~/fW (ffi) , 

x / , x" = a X — X 

x' and x" being regarded as independent. 

5. Form the increments of a function /for equicrescent values of the variable : 

A 1 f = f(a + h) -/(a), A,/ = /(a + 2A)-/(a + h), 

A s f = f{a + Zh)-f{a + 27t), ••-. 



FUNDAMENTAL THEOEY 4& 

These are called first differences ; the differences of these differences are 

Aff = f(a +2h)- 2f(a + k) +/(a), 

A|/ =f(a + 3 h) - 2f(a + 2 h) + f(a + h), ■ ■ ■ 

which are called the second differences ; in like manner there are third differences. 

A}f=f(a + 3 h) - Sf(a + 2h)+ Sf(a + h) -/(a), ■ ■ • 

and so on. Apply the Law of the Mean to all the differences and show that 

A?/ = h*f"(a + 6 x h + 2 h), A?f = h s f'"{a + x h + 6Ji + s h), ■ •■. 

Hence show that if the first n derivatives of / are continuous at a, then 

/"(a) = lim ^f, f'"(a) =lim^f, • • • , /(»)(a) = lim ^- . 

h = IV h=0 IV h = IV 1 

6. Cauchy's Theorem. If f(x) and <j>(x) are continuous over a Si x ==i 6, have- 
derivatives at each interior point, and if 0' (x) does not vanish in the interval, 

f(b)-f(a) = /'(g) Qr /(q + ft)-/( a ) = /(g + flfi.) _ 
(6) - (a) 0'(|) (a + h) - (a) 0'(a + #/*) ' 

Prove that this follows from the application of Rolle's Theorem to the function 

* (x) = f{x) - f{a) - [0 (x) - («)] { ( f~{f| • 

0(6) -0(a) 

7. One application of Ex. 6 is to the theory of indeterminate forms. Show that 
if /(a) = (a) = and if f'(x)/<p'(x) approaches a limit when x = a, then/(x)/0 (x) 
will approach the same limit. 

8. Taylor's Theorem. Xote that the form /(&) = /(«) + (o — a)/'(£) is one way 
of writing the Theorem of the Mean. By the application of Rolle's Theorem to 

, (x) =/(6 ) - /( x) - (ft - OT ) - (6 - s)» ^ "^ ~ ( & - g > ™ , 

(6 - a) 2 

show /(&) = /(a) + (6 - o)/'( tt ) + (& ~ CT)2 / // (g), 

and to * (x) = /(&) - f(x) - (6 - x)/'(x) - ^=-^/" (x) ^"^""V -« (x> 

2 (n-1)! 



(6 - a) 



^-•••-^/-4 



show /(&) = /(«) + (6 - «)/'(«) + (b 2 n) V "(«) + ■ ■ • 

(n — 1) ! ?i ! 

What are the restrictions that must be imposed on the function and its derivatives ?' 

9. If a continuous function over a^x^b has a right-hand derivative at each 
point of the interval which is zero, show that the function is constant. Apply Ex. 2 
to the functions f(x) + e (x — a) and/(x) — e(x — a) to show that the maximum 
difference between the functions is 2 e (6 — a) and that / must therefore be constant. 



50 



INTEODUCTOEY EEVIEW 



10. State and prove the theorems implied in the formulas (4)-(6), p. 2. 

11. Consider the extension of Ex. 7, p. 44, to derivatives of functions defined 
over a dense set. If the derivative exists and is uniformly continuous over the dense 
set, what of the existence and continuity of the derivative of the function when its 
definition is extended as there indicated ? 

12. If f(x) has a finite derivative at each point of the interval a^x^b, the 
derivative /'(») must take on every value intermediate between any two of its values. 
To show this, take first the case where f'(a) and/'(&) have opposite signs and show, 
by the continuity of / and by Theorem 13 and Ex. 2, that /'(£) = 0. Next if 
f'(a)<n<f'(b) without any restrictions on f\a) and/' (6), consider the function 
f(x) — fix and its derivative f\x) — /jl. Finally, prove the complete theorem. It 
should be noted that the continuity of f'(x) is not assumed, nor is it proved ; for 
there are functions which take every value intermediate between two given values 
and yet are not continuous. 

28. Summation and integration. Let f(x) be denned and limited 
over the interval a ^ x ^ b and let M, m, and = M — m be the 
upper frontier, lower fron- 
tier, and oscillation of fix) 
in the interval. Let n — 1 
points of division be intro- 
duced in the interval divid- 
ing it into n consecutive 
intervals S 1? S 2 , • • •, S n of 
which the largest has the 
length A and let M iy m i} O ti 

and f(i t ) be the upper and lower frontiers, the oscillation, and any 
value of the function in the interval S { . Then the inequalities 

m g. ^ m .8. =s /(£)8. =s M.%. ^ M$t 
will hold, and if these terms be summed up for all n intervals, 



Y 










Mi 


M 








s 


Mi V 
















m 















c 


c 


I 


>i 


I 


) X 



m 



(b - a) ^ 5)mA ^ ]£/(£) 8, =s Jj JfefA ^ M(b - a) (A) 



will also hold. Let s = Swrf, o- = V(£)A-, and s = 3M&. From (A) 
it is clear that the difference S — s does not exceed 

(M - m) (b - a) = 0(b - a), 

the product of the length of the interval by the oscillation in it. The 
values of the sums S, s, <r will evidently depend on the number of parts 
into which the interval is divided and on the way in which it is divided 
into that number of parts. 

Theorem 19. If n' additional points of division be introduced into 
the interval, the sum S' constructed for the n + n' — 1 points of division 



FUNDAMENTAL THEORY 51 

cannot be greater than S and cannot be less than S by more than 
n'OA. Similarly, s' cannot be less than s and cannot exceed s by more 
than n'OA. 

Theorem 20. There exists a lower frontier i for all possible methods 
of constructing the sum S and an upper frontier I for s. 

Theorem 21. Darboux's Theorem. When e is assigned it is possible 
to find a A so small that for all methods of division for which & t =s A, 
the sums S and s shall differ from their frontier values i and I by less 
than any preassigned e. 

To prove the first theorem note that although (A) is written for the whole inter- 
val from a to b and for the sums constructed on it, yet it applies equally to any 
part of the interval and to the sums constructed on that part. Hence if Si = Midi be 
the part of S due to the interval 5* and if S' { be the part of S' due to this interval 
after the introduction of some of the additional points into it, mrf; s= S' { =s Si = Midi. 
Hence S- is not greater than Si (and as this is true for each interval 5,-, S' is not 
greater than S) and, moreover, Si — S' { is not greater than OA- and a fortiori not 
greater than OA. As there are only n' new points, not more than n" of the intervals 
dt can be affected, and hence the total decrease S — S / in S cannot be more than 
?i'OA. The treatment of s is analogous. 

Inasmuch as (A) shows that the sums S and s are limited, it follows from Theo- 
rem 4 that they possess the frontiers required in Theorem 20. To prove Theorem 21 
note first that as L is a frontier for all the sums S, there is some particular sum S 
which differs from L by as little as desired, say | e. Tor this S let n be the number 
of divisions. Xow consider S' as any sum for which each 5, is less than A = \ e/nO. 
If the sum S" be constructed by adding the n points of division for S to the points 
of division for S', S" cannot be greater than S and hence cannot differ from L by 
so much as \ e. Also S" cannot be greater than S / and cannot be less than S / by 
more than nOA, which is \e. As S" differs from L by less than \e and S' differs 
from S" by less than \ e, S' cannot differ from L by more than e, which was to be 
proved. The treatment of s and I is analogous. 

29. If indices are introduced to indicate the interval for which the 
frontiers L and I are calculated and if f3 lies in the interval from a to b, 
then if and /f will be functions of (3. 

Theorem 22. The equations L% = L£ -f L h c , a<c<b; L* = — L£ ; 
L^ = fi(b — a), m ^ /jl ^ M, hold for L, and similar equations for I. As 
functions of /?, if and l£ are continuous, and if fix) is continuous, 
they are differentiable and have the common derivative ~f(&). 

To prove that i a & = i a c + i c 6 , consider c as one of the points of division of the 
interval from a to b. Then the sums S will satisfy S% — S a c + S*, and as the limit 
of a sum is the sum of the limits, the corresponding relation must hold for the 
frontier L. To show that L% = — i 6 a it is merely necessary to note that S% = — <S" 
because in passing from b to a the intervals 8{ must be taken with the sign opposite 
to that which they have when the direction is from a to b. From (^1) it appears 
that m (b — a) ^ S a & ^ M (b — a) and hence in the limit m (b — a) ^ L% ^ M (b — a). 



52 INTRODUCTORY REVIEW 

Hence there is a value /*, m^i/x^M, such that i a & = /*(&— a). To show that L& 
is a continuous function of /3, take K >\M\ and |ra|, and consider the relations 

L! + h - Li = L* + L$ + h - L£ = L? + h = M, \n\<K, 

LS~ h -L! = Lt h -L!~ h -L$_ h =-L$_ h =-»% \fi\<K. 

Hence if e is assigned, a 5 may be found, namely 5 < e/K, so that | L£ ~ h — L® | < e 
when h < d and L& is therefore continuous. Finally consider the quotients 

and 



h -h 

where /x is some number between the maximum and minimum of f(x) in the inter- 
val j8 ==§ x ==§ /3 + h and, if / is continuous, is some value /(£) of / in that interval 
and where f/ = /(£') is some value of / in the interval (3 — h^=x^== (3. Now let 
h = 0. As the function /is continuous, lim/(|) = /(/3) and lim/(^) = f(p). Hence 
the right-hand and left-hand derivatives exist and are equal and the function L& 
has the derivative f(fi). The treatment of I is analogous. 

Theorem 23. For a given interval and function /, the quantities I 
and L satisfy the relation I =s L ; and the necessary and sufficient con- 
dition that L — lis that there shall be some division of the interval 
which shall make 2 (M { — m t -) B { = S0A < e. 

If L% = la, the function f is said to be integrable over the interval 

from a to b and the integral I fix) dx is denned as the common value 

Ll = 1%. Thus the definite integral is defined. 

Theorem 24. If a function is integrable over an interval, it is inte- 
grable over any part of the interval and the equations 

Xc r*b r\b 

fix) dx -+- J f(x) dx = I fix) dx, 
k) c %J a 

J/-» b /-» a r*b 

fix) dx = — I fix) dx, I fix) dx — fi(b — a) 

a Ub %J a 

hold; moreover, J f(x)dx = F(/3) is a continuous function of ft; and 
\i fix) is continuous, the derivative F\f$) will exist and be f(fi). 

By [A) the sums S and s constructed for the same division of the interval satisfy 
the relation S — s === 0. By Darboux's Theorem the sums S and s will approach the 
values L and I when the divisions are indefinitely decreased. Hence L — I §e= 0. 
Now if L = I and a A be found so that when 5{ < A the inequalities S — L < | e and 
£•— s < | e hold, then S — s = 2 (ifj — m*) 5* = 20/5; < e ; and hence the condition 
20;5; < e is seen to be necessary. Conversely if there is any method of division such 
that 20*5* < e, then S — s < e and the lesser quantity X — I must also be less than e. 
But if the difference between two constant quantities can be made less than e, 
where e is arbitrarily assigned, the constant quantities are equal ; and hence the 



FUNDAMENTAL THEOEY 53 

condition is seen to be also sufficient. To show that if a function is integrable over 
an interval, it is integrable over any part of the interval, it is merely necessary to 
show that if L% = 1%, then Lj* = l£ where a and /3 are two points of the interval. 
Here the condition 20,<5,<6 applies; for if 20, 5; can be made less than e for the 
whole interval, its value for any part of the interval, being less than for the whole, 
must be less than e. The rest of Theorem 24 is a corollary of Theorem 22. 

30. Theorem 25. A function is integrable over the interval a ^ x ^ b. 
if it is continuous in that interval. 

Theorem 26. If the interval a ^ x ^ b over which fix) is defined 
and limited contains only a finite number of points at which / is dis- 
continuous or if it contains an infinite number of points at which / is 
discontinuous but these points have only a finite number of points of 
condensation, the function is integrable. 

Theorem 27. If /(;/:) is integrable over the interval a^x^b, the 

sum a = %f(fi) S, will approach the limit J f(x) dx when the indi- 

\J a 

vidual intervals 8,- approach the limit zero, it being immaterial how 

they approach that limit or how the points £ t - are selected in their 
respective intervals &V 

Theore3i 28. If fix) is continuous in an interval a ^ x === b, then 

fix) has an indefinite integral, namely ) f(x) dx, in the interval. 



Theorem 25 may be reduced to Theorem 23. For as the function is continuous, 
it is possible to find a A so small that the oscillation of the function in any interval 
of length A shall be as small as desired (Theorem 9). Suppose A be chosen so' that 
the oscillation is less than e/(b — a). Then 20,5; < e when 5; < A : and the function 
is integrable. To prove Theorem 26, take first the case of a finite number of discon- 
tinuities. Cut out the discontinuities surrounding each value of x at which /is dis- 
continuous by an interval of length 8. As the oscillation in each of these intervals 
is not greater than 0, the contribution of these intervals to the sum 20 2 -5j is not 
greater than Ond. where n is the number of the discontinuities. By taking 5 small 
enough this may be made as small as desired, say less than \ e. Xow in each of the 
remaining parts of the interval a = x =!== b. the function / is continuous and hence 
integrable, and consequently the value of 20fS, for these portions may be made as 
small as desired, say \ e. Thus the sum 20;5; for the whole interval can be made 
as small as desired and/(x) is integrable. When there are points of condensation 
they may be treated just as the isolated points of discontinuity were treated. After 
they have been surrounded by intervals, there will remain over only a finite num- 
ber of discontinuities. Further details will be left to the reader. 

For the proof of Theorem 27. appeal may be taken to the fundamental relation 
(A) which shows that s =s <? rs S. Xow let the number of divisions increase indefi- 
nitely and each division become indefinitely small. As the function is intestable, 
S and s approach the same limit I f(x)dx. and consequently <r # which is included 
between them must approach that limit. Theorem 28 is a corollary of Theorem 24 



54 INTEODUCTQEY REVIEW 

which states that as/(x) is continuous, the derivative of / fix) dx is/(x). By defi- 
nition, the indefinite integral is any function whose derivative is the integrand. 
f(x)dx is an indefinite integral of /(x), and any other may be obtained 

a 

by adding to this an arbitrary constant (Theorem 16). Thus it is seen that the 
proof of the existence of the indefinite integral for any given continuous function 
is made to depend on the theory 1 of definite integrals. 

EXERCISES 

1. Rework some of the proofs in the text with I replacing L. 

2. Show that the L obtained from C/(x), where C is a constant, is C times the L 
obtained from /. Also if it, u, w are all limited in the interval a =§ x ^ 6, the L for 
the combination u + v — w will be L (u) + L (v) — L(w), where L (u) denotes the L 
for it, etc. State and prove the corresponding theorems for definite integrals and 
hence the corresponding theorems for indefinite integrals. 

3. Show that 20^- can be made less than an assigned e in the case of the func- 
tion of Ex. 6, p. 44. Note that I = 0, and hence infer that the function is integrable 
and the integral is zero. The proof may be made to depend on the fact that there 
are only a finite number of values of the function greater than any assigned value. 

4. State with care and prove the results of Exs. 3 and 5, p. 29. What restric- 
tion is to be placed on/(x) if /(£) may replace /x ? 

5. State with care and prove the results of Ex. 4, p. 29, and Ex. 13, p. 30. 

6. If a function is limited in the interval a === x === b and never decreases, show 
that the function is integrable. This follows from the fact that 20 t - =s is finite. 

7. More generally, let /(a) be such a function that 20; remains less than some 
number K, no matter how the interval be divided. Show that/ is integrable. Such 
a function is called a function of limited variation (§ 127). 

8. Change of variable. Let f(x) be continuous over a =s x ±s.&. Change the 
variable to x = <f>(t), where it is supposed that a = 0(^) and b = 0(£ 2 ), and that 
(t), 0'(£), and/[0 (t)~\ are continuous in t over t r ^t^ t 2 . Show that 

f{x)dx= f[<t>(t)]<p'(t)dt or | f(x)dx= / f[<p(t)]<p'(t)dt. 

Do this by showing that the derivatives of the two sides of the last equation with 
respect to t exist and are equal over ^ == t === t 2 , that the two sides vanish when 
t = t x and are equal, and hence that they must be equal throughout the interval. 

9. Osgood's Theorem. Let an be a set of quantities which differ uniformly from 
f{ki) Si by an amount frSi, that is, suppose 

oct = / (&) 8i + tfi, where [ ft ] < e and a ^ £ ^ b. 
Prove that if /is integrable, the sum 2a 4 - approaches a limit when 5; = and that 

r b 

the limit of the sum is I f(x) dx. 

J a 

10. Apply Ex. 9 to the case Af = f'Ax + fAx where /' is continuous to show 
directly th&tf(b) r f(a) = f f'(x) dx. Also by regarding Ax = 0' (t) At + fA«, apply 
to Ex. 8 to prove the rule for change of variable. 



PART I. DIFFERENTIAL CALCULUS 

CHAPTER III 

TAYLOR'S FORMULA AND ALLIED TOPICS 

31. Taylor's Formula. The object of Taylor's Formula is to express 
the value of a function f(x) in terms of the values of the function and 
its derivatives at some one point x = a. Thus 

/(*) =/(«) + (* - «)/'(«) + (£ ^P-/» +■■■ 

+ ( (»~-jsr / "" i '< a)+a - w 

Such an expansion is necessarily true because the remainder R may be 
considered as defined by the equation ; the real significance of the 
formula must therefore lie in the possibility of finding a simple ex- 
pression for R, and there are several. 

Theorem. On the hypothesis that f(x) and its first n derivatives 
exist and are continuous over the interval a =§= x ^ b, the function may 
be expanded in that interval into a polynomial in x — a, 

/(*) =/(«) + 0* - «)/'(«) + ^yf 1 /"^) + ■■■ 

+ (n-l). /"-"(") + *. (1) 

with the remainder i? expressible in any one of the forms 

= (^3^ Jp"~ 1 ->' < " > ( a + *-'>*■ (2) 

where h = x — a and a<£<a;or£ = a+ 07? where < $ < 1. 

A first proof may be made to depend on Rolle's Theorem as indicated in Ex. 8, 
p. 49. Let x be regarded for the moment as constant, say equal to b. Construct 

55 



SQ DIFFERENTIAL CALCULUS 

the function \p (x) there indicated, Note that \f/ (a) = \p (b) = and that the deriva- 
tive ^'(x) is merely 

n*) = - {b r x) Z l fin) <*> + n V x) 7n \ f w - w -v- *)■») 

(n— 1)! (o— a) w L 

^ -*)-> -!) (a) 1. 

By Kolle's Theorem ^'(£) = 0. Hence if £ be substituted above, the result is 

f(b) =f(a) + (b - a)f\a) +■■■ + ( 5 ~ a >"~V »-i>(a) + &ZL£>!/c»> ($), 

(n — 1) ! n ! 

after striking out the factor — (b — £) w_1 , multiplying by (6 — a) n /n, and transposing 

/(&). The theorem is therefore proved with the first form of the remainder. This 

proof does not require the continuity of the nth derivative nor its existence at a and at b. 

The second form of the remainder may be found by applying Rolle's Theorem to 

* (x) =/<&) -f(x) - (b - xjf'{x) ^-^""V —D {X) - (b - X) P, 

{11 1) . 

where P is determined so that B = (6 — a) P. Note that ^ (&) = and that by 
Taylor's Formula \p (a) = 0. Now 

fix) =- (6 ~ ^""V 00 ^) + -P or P =/(»)({> ^~^ n " 1 since ^'(£) = 0. 
r w (n - 1) !• v ■■ V (>i - 1) ! w 

Hence if if be written % = a+6h where h=b—a, then b—% = b—a—6h=(b—a)(l—0). 

And B = (>-a)P = ( 6 -a) < ft - a >-- 1 <;- g >-V ->(0 = (ft -° ) " (1 -, ,>) -V >(t). 

(n - 1) ! {n - 1) ! 

The second form of R is thus found. In this work as before, the result is proved 
for x = b, the end point of the interval a =5 x === 6. But as the interval could be 
considered as terminating at any of its points, the proof clearly applies to any x 
in the interval. 

A second proof of Taylor's Formula, and the easiest to remember, consists in 
integrating the nth derivative n times from a to x. The successive results are 

f X f(n)( X )dX =fn-l(x)lf=f(n-l)( X ) _/(*-l)( a ). 

f X f 7 (n) (*) dx 2 = f 7(»-D (x) dx- f /(»-•!) (a) dx 

= /(» - 2 > (x) - /< M - 2 > (a) - (x - a) /(» - 1) (a) . 
J " J " J /(»? (x) dx 3 = /<» - 3) (x) _/<»-8) (a) - (x - a)/(»-2) (a) - fc^!/(»-i) (a). 

f X - ■ • J/W (x) dx- = /(») -/(a) - (x - a)/' (a) 

- <*-">>» ( g - fl )-V-D (a). 

2 1 w (n - 1) ! 

Jo a: /1 x 

• • • I /0)(x) dx n . To trans- 

form this to the ordinary form, the Law of the Mean may be applied ((65), § 16). For 

Jt x ( r< n\n r* x /» X (x — a) n 
/('O(x) dx <M(x- a), m * ^ < I • • • / /<»>(x) dx« < 2f i -S 



TAYLOR'S FORMULA ; ALLIED TOPICS 57 

where m is the least and M the greatest value of /(«)(.r) from a to x. There is then 
some intermediate value /<")(£) = ^ such that 



f-f 



'/U>(z) ** = &—■ ^7<">(£). 



This proof requires that the nth derivative be continuous and is less general. 

The third proof is obtained by applving successive integrations bv parts to the 

r h 
obvious identity /(a + h) —/(a) = I /'(a + h — t) dt to make the integrand contain 

«/o 
higher derivatives. 

f(a + h)- f(a) = J V(« + h-t)dt = tf'(a + A - *)1 * + J * £/"(a + h - t) dt 

= hf\a) + i * 2 /"(a + h-t)] h + f % £ 2 /'"(« + /< - *') c?£ 
Jo ^o 

= hf'(a) + ?/"(a) + • • • + t^-/ ( " - 1} («) + f *7^ Z Tr I /° ,) ( fl + *- *) ^ 

2 ! (w — 1)! ^o (n — 1)1 

This, however, is precisely Taylor's Formula with the third form of remainder. 

If the point a about which the function is expanded is x = 0, the 
expansion will take the form known as Maelaurin's Formula : 

/W=/(0)+ ^0) + ^/'X0)+... + ^£^^- 1 ) (0)+*, (3) 

nV y (n — 1)! y y (>z — l).J f ' 

32. Both Taylor's Formula and its special case, Maclaurin's, express 
a function as a polynomial in h = x — a, of which all the coefficients 
except the last are constants while the last is not constant but depends 
on h both explicitly and through the unknown fraction which itself is 
a function of h. If, however, the Tith derivative is continuous, the coeffi- 
cient f {n) (a + 6Ji)/n\ must remain finite, and if the form of the deriva- 
tive is known, it may be possible actually to assign limits between 
which f^ n \a + 07i)/n ! lies. This is of great importance in making 
approximate calculations as in Exs. 8 ff. below ; for it sets a limit to 
the value of R for any value of n. 

Theorem. There is only one possible expansion of a function into 
a polynomial in h = x— a of which all the coefficients except the last 
are constant and the last finite ; and hence if such an expansion is 
found in any manner, it must be Taylor's (or Maclaurin's). 

To prove this theorem consider two polynomials of the ?ith order 

c + c x h + c 2 /i 2 + • • • + c n _i/t»-i + c n h a = C + C x h + C. 2 h 2 + ■■-+ C n ^h n - 1 + C n h>\ 

which represent the same function and hence are equal for all values of h from 
to b — a. It follows that the coefficients must be equal. For let h approach 0. 



58 DIFFERENTIAL CALCULUS 

The terms containing h will approach and hence c and C may be made as 
nearly equal as desired ; and as they are constants, they must be equal. Strike 
them out from the equation and divide by h. The new equation must hold for all 
values of h from to b — a with the possible exception of 0. Again let h = and 
now it follows that c 1 = C v And so on, with all the coefficients. The two devel- 
opments are seen to be identical, and hence identical with Taylor's. 

To illustrate the application of the theorem, let it be required to find the expan- 
sion of tan x about when the expansions of sin x and cos x about are given. 

sin x = x — I x s + T *-o x 5 + Px 7 , cos x = 1 — \ x 2 + ^ £ 4 + Qx 6 , 

where P and Q remain finite in the neighborhood of x = 0. In the first place note 
that tan x clearly has an expansion ; for the function and its derivatives (which 
are combinations of tan x and sec x) are finite and continuous until x approaches ^ tt. 
By division, 

X + I X 3 + T 2 5 x5 

1 - f x 2 + J ¥ « 4 + Qx 6 )x - ix 3 + T |o x 5 i + Px 7 

X - -I X 3 + 2V X 5 j + QX 7 



1 X 3_ J 1_ X 5: + {P _ Q)X 7 
1X 3 - 1 X 5 : + T yx 7 + |QX 9 



T5 



Hence tan x = x+|x 3 + t 2 txM x 7 , where S is the remainder in the division 

cosx 
and is an expression containing P, Q, and powers of x ; it must remain finite if P 
and Q remain finite. The quotient *S/cos x which is the coefficient of x 7 therefore 
remains finite near x = 0, and the expression for tan x is the Maclaurin expansion 
up to terms of the sixth order, plus a remainder. 

In the case of functions compounded from simple functions of which the expan- 
sion is known, this method of obtaining the expansion by algebraic processes upon 
the known expansions treated as polynomials is generally shorter than to obtain 
the result by differentiation. The computation may be abridged by omitting the 
last terms and work such as follows the dotted line in the example above ; but if 
this is done, care must be exercised against carrying the algebraic operations too 
far or not far enough. In Ex. 5 below, the last terms should be put in and carried 
far enough to insure that the desired expansion has neither more nor fewer terms 
than the circumstances warrant. 

EXERCISES 

1. Assume R=(b- a)*P; show R = ^ (1 ~ ^"~* >° (£)■ 

(n — 1)! k 

2. Apply Ex. 5, p. 29, to compare the third form of remainder with the first. 

3. Obtain, by differentiation and substitution in (1), three nonvanishing terms : 

(a) sin-ix, a = 0, ((3) tanh x, a — 0, (7) tan x, a = ± tv, 

(5) cscx, a = i7r, (e) e sinx , a = 0, ({) log sinx, a = | it. 

4. Find the nth derivatives in the following cases and write the expansion : 

(a) sin x, a = 0, (/3) sin x, a = \ ?r, (7) c x , a = 0, 

(5) c*, a = 1, (e) logx, a = 1, (f) (1 + x)*, a = 0. 



TAYLOE'S FOEMULA; ALLIED TOPICS 59 

5. By algebraic processes find the Maclaurin expansion to the term in x 5 : 



(a) sec x, (p) tanh x, (7) — v 1 — x 2 , 

(5) e^sinx, (e) [log(l-x)] 2 , (f) + Vcosh x, 

(17) e sin3; , (#) logcosx, (t) log Vl + x 2 . 

The expansions needed in this work may be found by differentiation or taken 
from B. O. Peirce's "Tables." In (7) and (f) apply the binomial theorem of Ex. 
4 (f). In (77) let y = sin x, expand e^, and substitute for ?/ the expansion of sin x. 
In (0) let cos x = 1 — y. In all cases show that the coefficient of the term in x 6 
really remains finite when x = 0. 

6. If f(a + h) = c + c x h + c 2 /i 2 + • • • + Cn-ih 11 - 1 + c n h n , show that in 

f h f(a + h) dh = c Q h + ^ h* + ^h s + ■ ■ ■ + ^ h» + f h c n h»dh 
Jo 2 3 n Jo 

the last term may really be put in the form Ph n+1 with P finite. Apply Ex. 5, p. 29. 

p x dx 

7. Apply Ex. 6 to sin-i-x = | , etc., to find developments of 

J o Vl - x 2 

(a) sin - 1 x, (/3) tan- 1 x, (7) sinh -1 x, 

(5)logl±^, (e) f.V-cfa, (fl.r^*,fc. 

1 — x «/o «/o a; 

In all these cases the results may be found if desired to n terms. 

8. Show that the remainder in the Maclaurin development of e x is less than 
x n eV'i ! ; and hence that the error introduced by disregarding the remainder in com- 
puting e* is less than x n e x /n ! . How many terms will suffice to compute e to four 
decimals ? How many for e 5 and for e - 1 ? 

9. Show that the error introduced by disregarding the remainder in comput- 
ing log (1+ x) is not greater than x n /n if x > 0. How many terms are required for 
the computation of log 1£ to four places ? of log 1.2 ? Compute the latter. 

10. The hypotenuse of a triangle is 20 and one angle is 31°. Find the sides by 
expanding sin x and cos x about a = \tt as linear functions of x — I it. Examine 
the term in (x — £ it) 2 to find a maximum value to the error introduced by 
neglecting it. 

11. Compute to 6 places: (a) ei, (p) log 1.1, (7) sin 30', (5) cos 30'. During 
the computation one place more than the desired number should be carried along 
in the arithmetic work for safety. 

12. Show that the remainder for log (1 + x) is less than x n /n (1 + x) n if x < 0. 
Compute (a) log 0.9 to 5 places, (/3) log 0.8 to 4 places. 

13. Show that the remainder for tan- 1 x is less than x n /n where n may always 
be taken as odd. Compute to 4 places tan- 1 1. 

14. The relation 1 tt = tan- 1 1 = 4 tan- 1 \ — tan- 1 ^ enables \ ir to be found 
easily from the series for tan- 1 x. Find \ -k to 7 places (intermediate work carried 
to 8 places). 

15. Computation of logarithms, (a) If a = log Jg -, b = log f f , c = log f i, then 
log2 = 7a-26 + 3c, log3 = 11 a - 3b + 5c, log 5 = 16a - 46 + 7c. 



60 DIFFERENTIAL CALCULUS 

Now a — — log (1 — jL), b = — log (1 — T tjo)r c — l°o (1 + io) are readily computed 
and hence log 2, log 3, log 5 may be found. Carry the calculations of a, 6, c to 
10 places and deduce the logarithms of 2, 3, 5, 10, retaining only 8 places. Com- 
pare Peirce's "Tables,' 1 p. 109. 

1 + x 2 x n 

(8) Show that the error in the series for log is less than Com- 

v ' 1 - x 11(1- x)» 

pute log 2 corresponding to x = | to 4 places, log If to 5 places, log If to 6 places. 

g Lp + g 3 \p + g/ 2 n - 1 \p + g/ J 

give an estimate of fign+ij and compute to 10 figures log 3 and log 7 from log 2 
and log 5 of Peirce's "Tables " and from 

81 7 4 

4 log 3 — 4 log 2 — log 5 = log — , 4 log 7 — 5 log 2 — log 3 — 2 log 5 = log 



80 °7 4 -l 

16. Compute Ex. 7 (e) to 4 places for x = 1 and to 6 places for x = \. 

17. Compute sin- 1 0.1 to seconds and sin- 1 -! to minutes. 

18. Show that in the expansion of (1 + x) k the remainder, as x is > or < 0, is 

k- (k — 1)- ■ ■ (k — n) x n 



Rn< 



k.(k-l)...(k-n) xn 



1-2 



or R n < 



1-2---H (l+x)»-* 



, n>k. 



Hence compute to 5 figures Vl03, V98, V28, V250, VlOOO. 

19. Sometimes the remainder cannot be readily found but the terms of the 
expansion appear to be diminishing so rapidly that all after a certain point appear 
negligible. Thus use Peirce's "Tables," Nos. 774-789, to compute to four places 
(estimated) the values of tan 6°, log cos 10°, esc 3°, sec 2°. 

20. Find to within 1% the area under cos (x 2 ) and sin (x 2 ) from to \ it. 

21. A unit magnetic pole is placed at a distance L from the center of a magnet 
of pole strength M and length 2 Z, where l/L is small. Find the force on the pole 
if (a) the pole is in the line of the magnet and if (8) it is in the perpendicular 
bisector. 

4 Ml 1 1 \ 2 2 Ml 3 / 1 \ 2 

Arts, (a) — — (1 + e) with e about 2 ( - ) , (8) (1 — e) with e about - ( — ) • 

L 3 \LJ L° 2 \LJ 

22. The formula for the distance of the horizon is D =V|Zi where D is the 
distance in miles and h is the altitude of the observer in feet. Prove the formula 
and show that the error is about \% for heights up to a few miles. Take the radius 
of the earth as 3960 miles. 

23. Find an approximate formula for the dip of the horizon in minutes below 
the horizontal if h in feet is the height of the observer. 

24. If S is a circular arc and C its chord and c the chord of half the arc, prove 
8 = - 1 - (8 c - C) (1 + e) where e is about S 4 /7680 E 4 if R is the radius. 

25. If two quantities differ from each other by a small fraction e of their value, 
show that their geometric mean will differ from their arithmetic mean by about 
|e 2 of its value. 

26. The algebraic method may be applied to finding expansions of some func- 
tions which become infinite. (Thus if the series for cos x and sin x be divided to 
find cotx, the initial term is 1/x and becomes infinite at x = just as cotx does. 



TAYLOR'S FORMULA; ALLIED TOPICS 61 

Such expansions are not Maclaurin developments but are analogous to them. 
The function x cot x would, however, have a Maclaurin development and the 
expansion found for cotx is this development divided by x.) Find the develop- 
ments about x = to terms in x 4 for 

(a) cotx, (j8) cot 2 x, (7) cscx, (5) csc 3 x, 

(e) cot x cscx, (f) l/itan-ix) 2 , (77) (sinx — tanx)- 1 . 

27. Obtain the expansions : 
(a) log sinx = log x — \ x 2 — T |-o x 4 + i?, (/3) log tanx = logx + }x 2 + ¥ %x 4 + • • •, 
(7) likewise for log vers x. 

33. Indeterminate forms, infinitesimals, infinites. If two functions 
f(x) and <f> (x) are defined for x = a and if <f>(a)=£- 0, the quotient f/<j> is 
defined for x — a. But if <f> (a) = 0, the quotient f/<f> is not defined for a. 
If in this case / and <f> are defined and continuous in the neighborhood 
of a and f(a) =£ 0, the quotient will become infinite as x = a ; whereas 
if f(a) = 0, the behavior of the quotient f/<f> is not immediately appar- 
ent but gives rise to the indeterminate form 0/0. In like manner if / 
and <£ become infinite at a, the quotient f/<f> is not defined, as neither 
its numerator nor its denominator is defined ; thus arises the indeter- 
minate form cc/cc. The question of determining or evaluating an 
indeterminate form is merely the question of finding out whether the 
quotient f/<j> approaches a limit (and if so, what limit) or becomes 
positively or negatively infinite when x approaches a. 

Theorem. L' Hospital/ s Rule. If the functions /(a;) and <f>(x), which 
give rise to the indeterminate form 0/0 or oo/oo when x = a, are con- 
tinuous and differentiable in the interval a < x == b and if b can be 
taken so near to a that $(x) does not vanish in the interval and if the 
quotient /'/<£' of the derivatives approaches a limit or becomes posi- 
tively or negatively infinite as x = a, then the quotient f/(J> will ap- 
proach that limit or become positively or negatively infinite as the case 
may be. Hence an indeterminate form 0/0 or cc/x> may he replaced by 
the quotient of the derivatives of numerator and denominator. 

Case I. f(a) = <f> (a) = 0. The proof follows from Cauchy's Formula, Ex. 6, p. 49. 

For m = m-m = m, a<i<Zm 

0(x) <p(x)-<p(a) 0'(£) 

Now if x= a. so must £, which lies between x and a. Hence if the quotient on the 
right approaches a limit or becomes positively or negatively infinite, the same is 
true of that on the left. The necessity of inserting the restrictions that / and (p 
shall be continuous and differentiable and that <p' shall not have a root indefinitely 
near to a is apparent from the fact that Cauchy's Formula is proved only for func- 
tions that satisfy these conditions. If the derived form /'/(/>' should also be inde- 
terminate, the rule could again be applied and the quotient f" /<$>" would replace 
/'/<£' with the understanding that proper restrictions were satisfied by/', <j>', and 0". 



62 DIFFEBENTIAL CALCULUS 

Case II. f(a) = (a) = co. Apply Cauchy's Formula as follows : 

f(x)-f(b) = f(x) l-f(b)/f(x) == rg) > ' a<x<b, 

<p{x)-<f> (b) (x) 1 - (6)/0 (x) 0'(£) ' x < £ < 6, 

where the middle expression is merely a different way of writing the first. Now 
suppose that/'(x)/0'(x) approaches a limit when x = a. It must then be possible to 
take 6 so near to a that/ / (^)/0'(^) differs from that limit by as little as desired, no 
matter what value if may have between a and b. Now as / and become infinite 
when x = a, it is possible to take x so near to a that f{b)/f(x) and (6)/0 (x) are 
as near zero as desired. The second equation above then shows that /(x)/0 (x), 
multiplied by a quantity which differs from 1 by as little as desired, is equal to 
a quantity /'(£)/0'(£) which differs from the limit of f / {x)/4> / (x) as x = a by as little 
as desired. Hence //0 must approach the same limit as /'/0'. Similar reasoning 
would apply to the supposition that/ / /0 / became positively or negatively infinite, 
and the theorem is proved. It may be noted that, by Theorem 16 of § 27, the form 
/'/0' is sure to be indeterminate. The advantage of being able to differentiate 
therefore lies wholly in the possibility that the new form be more amenable to 
algebraic transformation than the old. 

The other indeterminate forms • co, 0°, 1°°, oo°, co — co may be reduced to the 
foregoing by various devices which may be indicated as follows : 

0-co = — = — , 0° = ei°so° = e 01 °g° = e° ,0 °, •••, co — co = loge 00 - 00 = log — . 
11 e 00 

co" 

The case where the variable becomes infinite instead of approaching a finite value 
a is covered in Ex. 1 below. The theory is therefore completed. 

Two methods which frequently may be used to shorten the work of evaluating 
an indeterminate form are the method of E -functions and the application of Taylor's 
Formula. By definition an E -function for the point x = a is any continuous function 
which approaches a finite limit other than when x = a. Suppose then that/(x) or 
0(x) or both may be written as the products E x f A and E 2 <p v Then the method of 
treating indeterminate forms need be applied only to/ 1 /0 1 and the result multiplied 
by lim E x /E 2 . For example, . 

lim = lim (x 2 + ax + a 2 ) lim = 3 a 2 lim == 3 a 2 . 

rr = a Sill (X — a) x = a x = a sin (x — a) x = a sill (x — O) 

Again, suppose that in the form 0/0 both numerator and denominator may be de- 
veloped about x — a by Taylor's Formula. The evaluation is immediate. Thus 

tan x — sin x _ (x + \ x 3 + Px 5 ) — (x — \ x 3 + Qx 5 ) _ \ + (P — Q) x 2 _ 
x 2 log(l + x) ~ x 2 (x - i x 2 + Bx s ) ~ 1 - i x + i?x 2 ' 

and now if x = 0, the limit is at once shown to be simply \. 

When the functions become infinite at x = a, the conditions requisite for Taylor's 
Formula are not present and there is no Taylor expansion. Nevertheless an expan- 
sion may sometimes be obtained by the algebraic method (§ 32) and may frequently 
be used to advantage. To illustrate, let it be required to evaluate cot x — 1/x which 
is of the form co — co when x = 0. Here 

cosx 1 + I x 2 + Px 4 1 1 - i x 2 + Px 4 1 r 1 
cot x = ' 1 



sin x x — \ x 3 + Qx 5 x 1 — \ x 2 + Qx 4 x 



(i -£* + *•) 



TAYLOR'S FORMULA; ALLIED TOPICS 63 

where S remains finite when x = 0. If this value be substituted for cot x, then 
lim ( cotx ) = lim ( x + Sx 3 ) = lhn ( x + Sx 3 ) = 0. 

x = 0\ X] x = 0\X 3 X) £C = 0\ 3 / 

34. An infinitesimal is a variable which is ultimately to approach the 
limit zero ; an infinite is a variable which is to become either positively 
or negatively infinite. Thus the increments Ay and Ax are finite quan- 
tities, but when they are to serve in the definition of a derivative they 
must ultimately approach zero and hence may be called infinitesimals. 
The form 0/0 represents the quotient of two infinitesimals ; # the form 
cc/oo, the quotient of two infinites ; and 0- oo, the product of an infin- 
itesimal by an infinite. If any infinitesimal a is chosen as the primary 
infinitesimal, a second infinitesimal ft is said to be of the same order as 
a if the limit of the quotient ft /a exists and is not zero when a = ; 
whereas if the quotient ft/a becomes zero, ft is said to be an infinites- 
imal of higher order than a, but of lower order if the quotient becomes 
infinite. If in particular the limit ft/a n exists and is not zero when 
a = 0, then ft is said to be of the nth order relative to a. The deter- 
mination of the order of one infinitesimal relative to another is there- 
fore essentially a problem in indeterminate forms. Similar definitions 
may be given in regard to infinites. 

Theorem. If the quotient ft/a of two infinitesimals approaches a 
limit or becomes infinite when aiO, the quotient ft' /a' of two infin- 
itesimals which differ respectively from ft and a by infinitesimals of 
higher order will approach the same limit or become infinite. 

Theorem. DuhameVs Theorem. If the sum %a t = q? -f- a 2 -\ }- a n 

of n positive infinitesimals approaches a limit when their number n 

becomes infinite, the sum Sft = ft l + ft 2 H h /?„, where each ft t differs 

uniformly from the corresponding a { by an infinitesimal of higher 
order, will approach the same limit. 

As a' — a is of higher order than a and /3' — /3 of higher order than /3, 

lim*l^ = o, lim^^ = or * = 1 + „, £=l+ft 

a p a (3 

where ?j and f are infinitesimals. Now a! = a (1 + 77) and /3' — /3 (1 + f). Hence 

— = — — — - and lim — = — , 
a' a 1 + 77 a' a 

provided j3/a approaches a limit ; whereas if j3/a becomes infinite, so will $' /a' . 
In a more complex fraction such as (/3 — 7) /a it is not permissible to replace /3 

* It cannot be emphasized too strongly that in the symbol 0/0 the 0's are merely sym- 
bolic for a mode of variation just as =c is; they are not actual 0's and some other nota- 
tion would be far preferable, likewise for • 00, 0°, etc. 



64 DIFFERENTIAL CALCULUS 

and 7 individually by infinitesimals of higher order ; for (3 — y may itself be of 
higher order than (3 or y. Thus tan x — sin x is an infinitesimal of the third order 
relative to x although tan x and sin x are only of the first order. To replace tan x 
and sin x by infinitesimals which differ from them by those of the second order or 
even of the third order would generally alter the limit of the ratio of tan x — sin x 
to x 3 when x = 0. 

To prove DuhamePs Theorem the /3's may be written in the form 

pi=a i (l+r H ), t = l, 2, -.., n, M<e, 

where the ^'s are infinitesimals and where all the ^'s simultaneously may be made 
less than the assigned e owing to the uniformity required in the theorem. Then 

| (/3 t + /3 2 + • • • + j8 n ) - (aq + a 2 + • • • + a n )\ = 1 Vl a x + V2 a 2 + • • • + Vn<x n \ < e2a. 

Hence the sum of the /3's may be made to differ from the sum of the <x's by less 
than eSa, a quantity as small as desired, and as 2a approaches a limit by hypoth- 
esis, so 2/3 must approach the same limit. The theorem may clearly be extended 
to the case where the ar's are not all positive provided the sum 2 1 ai\ of the abso- 
lute values of the a's approaches a limit. 

35. If y z=f(x), the differential of y is defined as 

dy =f'(x) Ax, and hence dx = 1 • Ax. (4) 

From this definition of dy and dx it appears that dy/dx =f'(x), where 
the quotient dy/dx is the quotient of two finite quantities of which, dx 
may be assigned at pleasure. This is true if x is the independent 
variable. If x and y are both expressed in terms of t, 

. x = x(t), y = y(t^ dx = Dpzdt, dy = D t ydt', 

and • g = g = £ x y, by virtue of (4), § 2. 

From this appears the important theorem : The quotient dy/dx is the 
derivative of y with respect to x no matter what the indep>endent variable 
may be. It is this theorem which really justifies writing the derivative 
as a fraction and treating the component differentials according to the 
rules of ordinary fractions. For higher derivatives this is not so, as 
may be seen by reference to Ex. 10. 

As Ay and Ax are regarded as infinitesimals in defining the deriva- 
tive, it is natural to. regard dy and dx as infinitesimals. The difference 
Ay — dy may be put in the form 



% - % = 



^-/'(*)]A*, (5) 



Ax 



wherein it appears that, when Ax = 0, the bracket approaches zero. 
Hence arises the theorem : If x is the inde]?endent variable and if Ay 
and dy are regarded as infinitesimals, the difference Ay — dy is an infin- 
itesimal of higher order than Ax. This has an application to the 



TAYLOK'S FORMULA; ALLIED TOPICS Qo 

subject of change of variable in a definite integral. For if x = <f>(t) f 
then dx = cf>'(t)dt, and apparently 

f f(x)dx= f % f£4>(t)-]4>'tyde, 

J a J ti 

where <f> (t x ) = a and <f> (t ) = b, so that t ranges from t x to t 2 when x 
ranges from a to b. 

But this substitution is too hasty ; for the dx written in the integrand 
is really Ax, which differs from dx by an infinitesimal of higher order 
when x is not the independent^ variable. The true condition may be 
seen by comparing the two sums 

^/(fc,) Ax„ ]£/[> ft)] *'&) At { , At = dt, 

the limits of which are the two integrals above. ]S"ow as Ax differs 
from dx=<j>'(t)dt by an infinitesimal of higher order, sof(x)Ax will 
differ from f [cf> (t)^\ <f>' (t) dt by an infinitesimal of higher order, and 
with the proper assumptions as to continuity the difference will be uni- 
form. Hence if the infinitesimals f(x) Ax be all positive, Duhaniel's 
Theorem may be applied to justify the formula for change of variable. 
To avoid the restriction to positive infinitesimals it is well to replace 
Duhaniel's Theorem by the new 

Theorem. Osgood's Theorem. Let a y a 2 , ••-, a n be n infinitesimals 
and let a { differ uniformly by infinitesimals of higher order than Ax 
from the elements f{x^)Ax i of the integrand of a definite integral 

Jf(x) dx, where / is continuous ; then the sum 2 a = a x + <r 2 -\ \-a n 
a 

approaches the value of the definite integral as a limit when the num- 
ber n becomes infinite. 

Let at =f(Xi)Aii + frAXi, where |ft| <e owing to the uniformity demanded. 



Then 



2)<*i - ^f( x d Ax i I = I ^tiAXi\<€^AXi = e{b - a). 



But as / is continuous, the definite integral exists and one can make 

,6 



dx 



<e(6-a+l). 



2^ f(*i) Aii — f f(x) dx I < e, and hence ^. a i — f f( x ) 

It therefore appears that 2a ( - may be made to differ from the integral by as little 
as desired, and Za t - must then approach the integral as a limit. Xow if this theo- 
rem be applied to the case of the change of variable and if it be assumed that 
f[<j>(t)~\ and <f>'(t) are continuous, the infinitesimals Axi and dX{ = <p' (U) dU will 
differ uniformly (compare Theorem 18 of § 27 and the above theorem on Ay — dy) 
by an infinitesimal of higher order, and so will the infinitesimals f(Xi) Axi and 
/[0 (U)] (p'(ti) dt t . Hence the change of variable suggested by the hasty substitution 
is justified. 



m DIFFERENTIAL CALCULUS 

EXERCISES 

1. Show that 1' Hospital's Rule applies to evaluating the indeterminate form 
f{x)/(p (x) when x becomes infinite and both/ and <j> either become zero or infinite. 

2. Evaluate the following forms by differentiation. Examine the quotients 
for left-hand and for right-hand approach ; sketch the graphs in the neighborhood 
of the points. 

, x , . & x — b* / „v i . tan x — 1 , x , . 

(a) Inn , (fi) Inn , (7) hm x log x, 

x = ' X x±\t; X — \ir a; = 

1 

(5) limxe-*, (e) lim (cotx) sina; , (f) limsc 1 -*. 

x=<x> x = x = l 

3. Evaluate the following forms by the method of expansions : 

(1 \ px _ pt&nx locrr 

- - cot 2 x ) , (|8) lim , (7) lim ^^ , 

x 2 / x = x — tan x * = 1 1 — x 

... .. . . . . . ,. x sin (sin x) — sin 2 x ,. e* — e-* — 2x 

(5) lim (cschx— cscx), (e) lim * f , (f) hm 

x = o x = o x b x = o x — sin x 

4 . Evaluate by any method : 

. . .. e^— e-* + 2sinx— 4x . . .. /tanx\^ 
(a) lim , (j8) lim , 

x = X° x = 0\ X / 

, . . . x cos 3 x — log (1 -1- x) — sin -14 x 2 .„, .. log Ix—lir) 

(7) hm & v ' ? — , (5) hm— ^ ?— i, 

ic = o x 3 x=1tt tanx 



(o 



iTl x ( i+ l) x - m *H 1+ M- 



5. Give definitions for order as applied to infinites, noting that higher order 
would mean becoming infinite to a greater degree just as it means becoming zero 
to a greater degree for infinitesimals. State and prove the theorem relative to quo- 
tients of infinites analogous to that given in the text for infinitesimals. State and 
prove an analogous theorem for the product of an infinitesimal and infinite. 

6. Note that if the quotient of two infinites has the limit 1, the difference of 
the infinites is an infinite of lower order. Apply this to the proof of the resolution 
in partial fractions of the quotient f(x)/F(x) of two polynomials in case the roots 
of the denominator are all real. For if F(x) = (x — a) k F 1 (x), the quotient is an 
infinite of order k in the neighborhood of x = a ; but the difference of the quotient 
and/(a)/(x — a) k F l (a) will be of lower integral order — and so on. 

7. Show that when x = +co, the function e x is an infinite of higher order 
than x n no matter how large n. Hence show that if P(x) is any polynomial, 
lim P (x) e~ x — when x = + 00. 

X= (X) 

8. Show that (log x) m when x is infinite is a weaker infinite than x n no matter 
how large m or how small n, supposed positive, may be. What is the graphical 
interpretation ? 

9. If P is a polynomial, show that lim P(-)e x2 = 0. Hence show that the 

x = \ x / 

-1 -- x» 

Maclaurin development of e x2 is/(x) = e x - = —f^(6x) if /(0) is defined as 0. 

n ! 



TAYLOR'S FORMULA; ALLIED TOPICS 67 

10. The higher differentials are defined as d n y =/(«)(x) (dxj n where x is taken 
as the independent variable. Show that d k x = for k > 1 if x is the independent 
variable. Show that the higher derivatives D*y, B*y. • • • are not the quotients 
d-y/dx-, d 3 y/dx s , ■ • • if x and y are expressed in terms of a third variable, but that 
the relations are 

o _ dhjdx — d-xdy 8 _ dx {dxdhj — dyd z x) — 3 d-x {dxd-y — dyd 2 x) 

pi/ ~ d? ' xV ~ dx~> ' "'" 

The fact that the quotient d n y/dx n , n > 1, is not the derivative when x and y are 
expressed parametrically militates against the usefulness of the higher differentials 
and emphasizes the advantage of working with derivatives. The notation d n y/dx n 
is, however, used for the derivative. Nevertheless, as indicated in Exs. 16-19, 
higher differentials may be used if proper care is exercised. 

11. Compare the conception of higher differentials with the work of Ex. 5, p. 48. 

12. Show that in a circle the difference between an infinitesimal arc and its 
chord is of the third order relative to either arc or chord. 

13. Show that if /3 is of the nth order with respect to a. and y is of the first 
order with respect to a, then /3 is of the nth order with respect to y. 

14. Show that the order of a product of infinitesimals is equal to the sum of the 
orders of the infinitesimals when all are referred to the same primary infinitesimal 
a. Infer that in a product each infinitesimal may be replaced by one which differs 
from it by an infinitesimal of higher order than it without affecting the order of the 
product. 

15. Let A and B be two points of a unit circle and let the angle A OB subtended 
at the center be the primary infinitesimal. Let the tangents at A and B meet at 
T. and OT cut the chord AB in 31 and the arc A B in C. Find the trigonometric 
expression for the infinitesimal difference TC — CM and determine its order. 

16. Compute d- (x sin x) = (2 cos x — x sin x) dx 2 -f (sin x + x cos x) d' 2 x by taking 
the differential of the differential. Thus find the second derivative of x sinx if.x is 
the independent variable and the second derivative with respect to t if x = 1 + t 2 . 

17. Compute the first, second, and third differentials, d-x ^ 0. 



(a) x 2 cosj, (/3) Vl — x log (1 — x), (y) xe 2x mix. 

18. In Ex. 10 take y as the independent variable and hence express B 2 y. Df.y 
in terms of D^x, B 2 x. Cf. Ex. 10, p. 11. 

19. Make the changes of variable in Exs. 8, 9, 12, p. 14. by the method of 
differentials, that is, by replacing the derivatives by the corresponding differential 
expressions where x is not assumed as independent variable and by replacing these 
differentials by their values in terms of the new variables where the higher differ- 
entials of the new independent variable are set equal to 0. 

20. Reconsider some of the exercises at the end of Chap. I, say, 17-19, 22, 23, 
27, from the point of view of Osgood's Theorem instead of the Theorem of the Mean. 

21. Find the areas of the bounding surfaces of the solids of Ex. 11, p. 18. 



68 DIFFERENTIAL CALCULUS 

22. Assume the law F=kmm / /r 2 of attraction between particles. Find the 
attraction of : 

(a) a circular wire of radius a and of mass M on a particle m at a distance r from 
the center of the wire along a perpendicular to its plane ; Ans. JcMmr (a 2 + r 2 )~*. 

(/3) a circular disk, etc., as in (a) ; Ans. 2kMma~ 2 (l — r/^r 2 + a 2 ). 

(7) a semicircular wire on a particle at its center ; Ans. ZkMm/ira 2 . 

(5) a finite rod upon a particle not in the line of the rod. The answer should 
be expressed in terms of the angle the rod subtends at the particle. 

(e) two parallel equal rods, forming the opposite sides of a rectangle, on each 
other. 

23. Compare the method of derivatives (§ 7), the method of the Theorem of the 
Mean (§ 17), and the method of infinitesimals above as applied to obtaining the for- 
mulas for (a) area in polar coordinates, (/3) mass of a rod of variable density, (7) pres- 
sure on a vertical submerged bulkhead, (5) attraction of a rod on a particle. Obtain 
the results by each method and state which method seems preferable for each case. 

24. Is the substitution dx = (p'{t)dt in the indefinite integral / fix) dx to obtain 
the indefinite integral ( f[<f> (<)] <f>'(t) dt justifiable immediately ? 

36. Infinitesimal analysis. To work rapidly in the applications of 
calculus to problems in geometry and physics and to follow readily the 
books written on those subjects, it is necessary to have some familiarity 
with working directly with infinitesimals. It is possible by making use 
of the Theorem of the Mean and allied theorems to retain in every ex- 
pression its complete exact value ; but if that expression is an infini- 
tesimal which is ultimately to enter into a quotient or a limit of a sum, 
any infinitesimal which is of higher order than that which is ultimately 
kept will not influence the result and may be discarded at any stage of 
the work if the work may thereby be simplified. A few theorems 
worked through by the infinitesimal method will serve partly to show 
how the method is used and partly to establish results which may be 
of use in further work. The theorems which will be chosen are : 

1. The increment Ax and the differential dx of a variable differ by 
an infinitesimal of higher order than either. 

2. If a tangent is drawn to a curve, the perpendicular from the curve 
to the tangent is of higher order than the distance from the foot of the 
perpendicular to the point of tangency. 

3. An infinitesimal arc differs from its chord by an infinitesimal of 
higher order relative to the arc. 

4. If one angle of a triangle, none of whose angles are infinitesimal, 
differs infinitesimally from a right angle and if h is the side opposite 
and if <j> is another angle of the triangle, then the side opposite <f> is 
h sin <£ except for an infinitesimal of the second order and the adjacent 
side is h cos <f> except for an infinitesimal of the first order. 



TAYLOE'S FORMULA; ALLIED TOPICS 



The first of these theorems has been proved in § 35. The second follows from 
it and from the idea of tangency. For take the cc-axis coincident with the tangent 
or parallel to it. Then the perpendicular is Ay and the distance from its foot to the 
point of tangency is Ax. The quotient Ay /Ax approaches as its limit because the 
tangent is horizontal ; and the theorem is proved. The theorem would remain true 
if the perpendicular were replaced by a line making a constant angle with the tangent 
and the distance from the point of tangency to the foot of the perpendicular were re- 
placed by the distance to the foot of the oblique line. For if Z PMN = 9, 

P/ 



PM 
TM 



PN esc e 



PN 



TN - PN cot 9 TN 



esc 9 



PX 
~TN 



cot 9 



MX 



and therefore when P approaches T with 9 constant, PM/ TM approaches zero and 
PM is of higher order than TM. 

The third theorem follows without difficulty from the assumption or theorem 
that the arc has a length intermediate between that of the chord and that of the 
sum of the two tangents at the ends of the chord. Let 1 and 9 2 be the angles 
between the chord and the tangents. Then 



AB AT+ TB-AB _ AM (sec 9 X - 1) + MB (sec 9 2 - 1) 



AM + MB 



AM+ MB 



AM + MB 



(6) 



& 




M 



coefficients remain necessarily finite. Hence the difference between the arc and 

the chord is an infinitesimal of higher order than the chord. As 

the arc and chord are therefore of the same order, the difference 

is of higher order than the arc. This result enables one to replace 

the arc by its chord and vice versa in discussing infinitesimals of 

the first order, and for such purposes to consider an infinitesimal 

arc as straight. In discussing infinitesimals of the second order, this substitution 

would not be permissible except in view of the further theorem given below in 

§ 37. and even then the substitution will hold only as far as the lengths of arcs are 

concerned and not in regard to directions. 

For the fourth theorem let 9 be the angle by which C departs from 90° and with 
the perpendicular BM as radius strike an arc cutting BC. Then by trigonometry 

AC = AM + MC = h cos + BM tan 0, 
BC = h sin0 + £Jf (sec 0-1). 

Now tan 9 is an infinitesimal of the first order with respect to 9 ; 
for its Maclaurin development begins with 9. And sec 9 — 1 
is an infinitesimal of the second order; for its development 
begins with a term in 9 2 . The theorem is therefore proved. 
This theorem is frequently applied to infinitesimal triangles, 
that is, triangles in which h is to approach 0. 

37- As a further discussion of the third theorem it may be recalled that by defi- 
nition the length of the arc of a curve is the limit of the length of an inscribed 
polygon, namely, 

s = lim (VAxf + Ay'l + Vax| + Ay; + 




MC 



+ Vaz w 2 + A2/ w 2 ), 



70 DIFFERENTIAL CALCULUS 

NOW ^/X^T^-^/^^= Ax* + Ay* -d x>-dy* 

VAx 2 + Ay 2 + Vdx 2 + dy 2 
_ (Ax - dx) (Ax + dx) + (Ay - dy) (Ay + dy) 
VAx 2 + Ay 2 + Vdx 2 + dy" 2 
VAx 2 + Ay 2 — Vdx 2 + dy 2 _ (Ax — dx) Ax + dx 



VAx 2 + Ay 2 VAx 2 + Ay 2 VAx 2 + Ay 2 + Vdx 2 + dy 2 

(Ay — dy) Ay + dy 

VAx 2 + Ay 2 VAx 2 + Ay 2 + Vdx 2 + dy 2 

But Ax — dx and A?/ — dy are infinitesimals of higher order than A x and Ay. 
Hence the righ t-hand sid e must approach zero as its limit and hence VAx 2 + Ay 2 
differs from Vdx 2 + dy 2 by an infinitesimal of higher order and may replace it in 
the sum 

s = lim ^V ^ Ax? + Ayf = lim ^> Vdx 2 + dy 2 = ( Vl + y' 2 dx. 

The length of the arc measured from a fixed point to a variable point is a func- 
tion of the upper limit and the differential of arc is 

ds = d f Vl + y /2 dx = Vl + y' 2 dx = Vdx 2 + dy 2 . 

To find the order of the difference between the arc and its chord let the origin 
be taken at the initial point and the x-axis tangent to the curve at that point. 
rThe expansion of the arc by Maclaurin's Formula gives 

s(x) = s(0) + xs'(O) + i xV(0) + \x*8T'(6x), 



yy 



o. 



where s(0) = 0, s'(O) = Vl + y /2 | = 1, s"(0) 

Vl + y' 2 

Owing to the choice of axes, the expansion of the curve reduces to 

y = f(x) = y (0) + xi/(0) + 1 x 2 y"(0x) = | xV'(fo), 

and hence the chord of the curve is 

c(x) =Vx 2 + y 2 = x Vl + \x\y" (Ox)]* = x (1 + x 2 P), 

where P is a complicated expression arising in the expansion of the radical by 
Maclaurin's Formula. The difference 

s(x)-c (x) = [x + ix 3 s"'(0x)] - [x (1 + x 2 P)] = x 3 (i s w (0x) - P). 

This is an infinitesimal of at least the third order relative to x. Now as both s (x) 
and c (x) are of the first order relative to x, it follows that the difference s (x) — c (x) 
must also be of the third order relative to either s (x) or c (x) . Note that the proof 
assumes that y" is finite at the point considered. This result, which has been 
found analytically, follows more simply though perhaps less rigorously from the 
fact that sec 6 X — 1 and sec 2 — 1 in (6) are infinitesimals of the second order with 
B x and 2 . 

38. The theory of contact of plane curves may be treated by means 
of Taylor's Formula and stated in terms of infinitesimals. Let two 
curves y = f(x) and y = g(x) be tangent at a given point and let the 



TAYLOR'S FORMULA; ALLIED TOPICS 71 

origin be chosen at that point with the a?-axis tangent to the curves.. 
The Maclaurin developments are 

y = f(*) = l/"(0K + • • • + 7^-rv; ^- 1 / fr - 1> (0) + h x(n) f (n) (°) + • • ■ 



2 J x ' ' (» — 1)! 



/i 



2/ = y (*) = | ?"(0) x 2 + ■ ■ • + -^— *»-y«-'>(0) + ^ ^«(0) + ■ ■ • . 

If these developments agree up to but not including the term in x n , the 
difference between the ordinates of the curves is 

/(as) - g (x) = i * [/ W(0) - /*>(<>)] + ■ ■ ■ , /«(0) # f»(0), 

and is an infinitesimal of the nth order with respect to x. The curves- 
are then said to have contact of order n — 1 at their point of tangency. 
In general when two curves are tangent, the derivatives /"(0) and #"(0) 
are unequal and the curves have simple contact or contact of the first 
order. 

The problem may be stated differently. Let PM be a line which 
makes a constant angle 6 with the cc-axis. Then, when P approaches T,. 
if RQ be regarded as straight, the proportion 

lim (PR : PQ) = lim (sin Z PQR : sin Z PRQ) = sin : 1 

shows that PR and PQ are of the same order. Clearly also the lines 
TM and TN are of the same order. Hence if 

PR PQ A 

lim ^, TN =£ 0, oo, then lim- — =£ 0, oo . / s 

(tny ' ' (rjf)- ' yyr^ 

Hence if two curves have contact of the (n — 1) st < ^t < ^Z^m! n 
order, the segment of a line intercepted between "*j£ 

the two curves is of the nth order with respect to 
the distance from the point of tangency to its foot. It would also be- 
of the nth. order with respect to the perpendicular TF from the point 
of tangency to the line. 

In view of these results it is not necessary to assume that the two 
curves have a special relation to the axis. Let two curves y = f(x) and 
y = g(x) intersect when x = a, and assume that the tangents at that point 
are not parallel to the y-axis. Then 

y = y, + (x - «)/'(«) + • . . + ^l^p /fr-'W + ^r-/ W («) + • • ■ 
y = y + (x - a) g\a) +... + £L=fg^/»-U( a ) + ^L ^ (a) + . . . 



72 DIFFERENTIAL CALCULUS 

will be the Taylor developments of the two curves. If the difference 
of the ordinates for equal values of x is to be an infinitesimal of the 
nth order with respect to x — a which is the perpendicular from the 
point of tangency to the ordinate, then the Taylor developments must 
agree up to but not including the terms in x n . This is the condition for 
contact of order n—1. 

As the difference between the ordinates is 

f(x) -g(x)=±(x- a)> [/<»>(«) - </<">(«)] + • • • , 

the difference will change sign or keep its sign when x passes through 
a according as n is odd or even, because for values sufficiently near to 
x the higher terms may be neglected. Hence the curves will cross each 
other if the order of contact is even, but will not cross each other if the 
order of contact is odd. If the values of the ordinates are equated to find 
the points of intersection of the two curves, the result is 

and shows that x = a is a root of multiplicity n. Hence it is said that 
two curves have in common as many coincident points as the order of 
their contact plus one. This fact is usually stated more graphically 
by saying that the curves have n consecutive points in common. It may 
be remarked that what Taylor's development carried to n terms does, is 
to give a polynomial which has contact of order n — 1 with the function 
that is developed by it. 

As a problem on contact consider the determination of the circle which shall 
have contact of the second order with a curve at a given point (a, y ) . Let 

y = y + (x - a)f'(a) + |(x - a) 2 /'» + ■ • ■ 

be the development of the curve and let y' —f\a) = tanr be the slope. If the 
circle is to have contact with the curve, its center must be at some point of the 
normal. Then if B denotes the assumed radius, the equation of the circle may be 
written as 

(x - a) 2 + 2 B sin r (x - a) + (y - y f -2B cos r (y - y ) = 0, 

where it remains to determine B so that the development of the circle will coincide 
with that of the curve as far as written. Differentiate the equation of the circle. 

dy B sin r + (x - a) (dy\ 

— = -, ( — ) = tanr =f'(a), 

dx B cos r — (y — y ) \dxj a ^ Jo 

d*y = [B cos r - (y - y )Y + [B sin t + (a - a)] 2 ^ A%\ 



dx 2 [B cos r — (y — y )] s \dx 2 / aj v B cos 3 r 



and y = y + (x— a)f'{a) + i (x — a) 



B COS 3 T 



TAYLOB'S FORMULA: ALLIED TOPICS 73 



is the development of the circle. The equation of the coefficients of (x — a) 

1 f „, x • v s"ec»T {1 + [/'(a)] 2 } 1 

— /"(a), gives E - 



Ecos 3 t " /"(a) /"(a) 

This is the well known formula for the radius of curvature and shows that the cir- 
cle of curvature has contact of at least the second order with the curve. The circle 
is sometimes called the osculating circle instead of the circle of curvature. 

39. Three theorems, one in geometry and two in kinematics, will 
now be proved to illustrate the direct application of the infinitesimal 
methods to such problems. The choice will be : 

1. The tangent to the ellipse is equally inclined to the focal radii 
drawn to the point of contact. 

2. The displacement of any rigid body in a plane may be regarded 
at any instant as a rotation through an infinitesimal angle about some 
point unless the body is moving parallel to itself. 

3. The motion of a rigid body in a plane may be regarded as the 
rolling of one curve upon another. 

For the first problem consider a secant PP' which may be converted into a 
tangent TT' by letting the two points approach until they coincide. Draw the 
focal radii to P and P' and strike arcs with F and F / as 
centers. As F'P + PF = F'P' + P'F = 2 a, it follows 
that XP = MP'. Now consider the two triangles PP'M 
and P'PN nearly right-angled at M and JSf. The sides 
PP', PM, PN, P*M, P'N are all infinitesimals of the 
same order and of the same order as the angles at F and 
F'. By proposition 4 of § 36 

MP* = PP' cos Z PP'M + e 17 NP = PP' cos Z P'PN + e 2 , 

where e x and e 2 are infinitesimals relative to MP' and NP or PP'. Therefore 

lim [cos Z PP'M - cos Z P'PNl = cos Z TPF' - cos Z T'PF = lim 6l ~ ^ = 0, 

J pp , 

and the two angles TPF' and T'PF are proved to be equal as desired. 

To prove the second theorem note first that if a body is rigid, its position is com- 
pletely determined when the position AB of any rectilinear segment of the body 
is known. Let the points A and B of the body be de- 
scribing curves A A' and BB' so that, in an infinitesimal 
interval of time, the line AB takes the neighboring posi- 
tion A 'B'. Erect the perpendicular bisectors of the lines \\\ ^'^^^Z^-^-^' 
AA' and BB' and let them intersect at 0. Then the tri- 
angles A OB and A' OB' have the three sides of the one 
equal to the three sides of the other and are equal, and 
the second may be obtained from the first by a mere rotation about through the 
angle A OA'= BOB'. Except for infinitesimals of higher order, the magnitude of 
the angle is AA'/OA or BB'/OB. Next let the interval of time approach so that 
A' approaches A and B' approaches B. The perpendicular bisectors will approach 





74 DIFFERENTIAL CALCULUS 

the normals to the arcs AA' and. BB' at A and £, and the point will approach 
the intersection of those normals. 

The theorem may then be stated that : At any instant of time the motion of a 
rigid body in a plane may be considered as a rotation through an infinitesimal angle 
about the intersection of the normals to the paths of any two of its points at that in- 
stant ; the amount of the rotation will be the distance ds that any point moves divided 
by the distance of that point from the instantaneous center of rotation ; the angular 
velocity about the instantaneous center will be this amount of rotation divided by the 
interval of time dt, that is, it will be v/r, where v is the velocity of any point of the body 
and r is its distance from the instantaneous center of rotation. It is therefore seen 
that not only is the desired theorem proved, but numerous other details are found. 
As has been stated, the point about which the body is rotating at a given instant 
is called the instantaneous center for that instant. 

As time goes on, the position of the instantaneous center will generally change. 
If at each instant of time the position of the center is marked on the moving plane 
or body, there results a locus which is called the moving centrode or body centrode ; 
if at each instant the position of the center is also marked on a fixed plane over 
which the moving plane may be considered to glide, there results another locus which 
is called the fixed centrode or the space centrode. From these definitions it follows 
that at each instant of time the body centrode and the space centrode intersect at 
the instantaneous center for that instant. Consider a series of 
positions of the instantaneous center as P_ 2 P_iPP 1 P 2 marked 
in space and Q_2Q-iQQiQ 2 marked in the body. At a given 
instant two of the points, say P and Q, coincide ; an instant 
later the body will have moved so as to bring Q x into coin- 
cidence with P x ; at an earlier instant Q_i was coincident with 
P-\. Now as the motion at the instant when P and Q are together is one of 
rotation through an infinitesimal angle about that point, the angle between PP X 
and QQ X is infinitesimal and the lengths PP X and QQ X are equal ; for it is by the 
rotation about P and Q that Q x is to be brought into coincidence with P v Hence 
it follows 1° that the two centrodes are tangent and 2° that the distances PP X = QQ X 
which the point of contact moves along the two curves during an infinitesimal inter- 
val of time are the same, and this means that the two curves roll on one another 
without slipping — because the very idea of slipping implies that the point of con- 
tact of the two curves should move by different amounts along the two curves, 
the difference in the amounts being the amount of the slip. The third theorem 
is therefore proved. 

EXERCISES 

1. If a finite parallelogram is nearly rectangled, what is the order of infinites- 
imals neglected by taking the area as the product of the two sides ? What if the 
figure were an isosceles trapezoid ? What if it were any rectilinear quadrilateral 
all of whose angles differ from right angles by infinitesimals of the same order ? 

2. On a sphere of radius r the area of the zone between the parallels of latitude 
X and X + d\ is taken as 2 irr cos X • rd\ the perimeter of the base times the slant 
height. Of what order relative to d\ is the infinitesimal neglected ? What if the 
perimeter of the middle latitude were taken so that 2 irr 1 cos (X + \ d\) d\ were 
assumed ? 




TAYLOR'S FORMULA; ALLIED TOPICS 75 

3. What is the order of the infinitesimal neglected in taking 4:irr 2 dr as the 
volume of a hollow sphere of interior radius r and thickness dr ? What if the mean 
radius were taken instead of the interior radius ? Would any particular radius be 
best ? 

4. Discuss the length of a space curve y = /(x), z = g(x) analytically as the 
length of the plane curve was discussed in the text. 

5. Discuss proposition 2, p. 68, by Maclaurin's Formula and in particular show 
that if the second derivative is continuous at the point of tangency, the infinites- 
imal in question is of the second order at least. How about the case of the tractrix 



a. a — Va 2 — x- r-z =■ 

y = - log + v a- - x 2 , 

2 a + v a 2 — x 2 

and its tangent at the vertex x = a ? How about s (x) — c (x) of § 37 ? 

6. Show that if two curves have contact of order n — 1, their derivatives will 
have contact of order n — 2. What is the order of contact of the kth derivatives 
fc < n - 1 ? 

7. State the conditions for maxima, minima, and points of inflection in the 
neighborhood of a point where /00(a) is the first derivative that does not vanish. 

8. Determine the order of contact of these curves at their intersections: 

V2 (x 2 + y 2 + z) = 3 (x + y) r 2 = a 2 cos2<p x 2 + y 2 = y 

W 5x 2 _ 6x?/ + 5y 2 = 8, KP) y 2 = f a(a- x), W x 3 + ?/ 3 = x?y. 

9. Show that at points where the radius of curvature is a maximum or mini- 
mum the contact of the osculating circle with the curve must be of at least the 
third order and must always be of odd order. 

10. Let PN be a normal to a curve and T'N a neighboring normal. If is the 
center of the osculating circle at P, show with the aid of Ex. 6 that ordinarily the 
perpendicular from to P'N is of the second order relative to the arc PP' and that 
the distance OX is of the first order. Hence interpret the statement : Consecutive 
normals to a curve meet at the center of the osculating circle. 

11. Does the osculating circle cross the curve at the point of osculation ? Will 
the osculating circles at neighboring points of the curve intersect in real points ? 

• 12. In the hyperbola the focal radii drawn to any point make equal angles with 
the tangent. Prove this and state and prove the corresponding theorem for the 
parabola. 

13. Given an infinitesimal arc AB cut at C by the perpendicular bisector of its 
chord AB. What is the order of the difference AC — BC ? 

14. Of what order is the area of the segment included between an infinitesimal 
arc and its chord compared with the square on the chord ? 

15. Two sides AB, AC of a triangle are finite and differ infinitesimally ; the 
angle at A is an infinitesimal of the same order and the side BC is either recti- 
linear or curvilinear. What is the order of the neglected infinitesimal if the area 
is assumed as \ AB 2 6 ? What if the assumption is J AB • AC • 0? 



76 DIFFERENTIAL CALCULUS 

16. A cycloid is the locus of a fixed point upon a circumference which rolls on 
a straight line. Show that the tangent and normal to the cycloid pass through the 
highest and lowest points of the rolling circle at each of its instantaneous positions. 

17. Show that the increment of arc As in the cycloid differs from 2 a sin |#d!i9 
by an infinitesimal of higher order and that the increment of area (between two 
consecutive normals) differs from 3 a 2 sin 2 i OdO by an infinitesimal of higher order. 
Hence show that the total length and area are 8 a and 37ra 2 . Here a is the radius 
of the generating circle and 6 is the angle subtended at the center by the lowest 
point and the fixed point which traces the cycloid. 

18. Show that the radius of curvature of the cycloid is bisected at the lowest 
point of the generating circle and hence is 4 a sin \ 6. 

19. A triangle ABC is circumscribed about any oval curve. Show that if the 
side BC is bisected at the point of contact, the area of the triangle will be changed 
by an infinitesimal of the second order when BC is replaced by a neighboring tan- 
gent B'C, but that if BC be not bisected, the change will be of the first order. 
Hence infer that the minimum triangle circumscribed about an oval will have its 
three sides bisected at the points of contact. 

20. If a string is wrapped about a circle of radius a and then unwound so that 
its end describes a curve, show that the length of the curve and the area between 
the curve, the circle, and the string are 

s= f a6d9, A = C \ aW&d, 

where 6 is the angle that the unwinding string has turned through. 

21. Show that the motion in space of a rigid body one point of which is fixed 
may be regarded as an instantaneous rotation about some axis through the given 
point. To do this examine the displacements of a unit sphere surrounding the fixed 
point as center. 

22. Suppose a fluid of variable density D(x) is flowing at a given instant through 
a tube surrounding the x-axis. Let the velocity of the fluid be a function v(x) of x. 
Show that during the infinitesimal time 8t the diminution of the amount of the 
fluid which lies between x = a and x = a + h is 

S [v (a + h) D{a + h) 8t-v (a) D {a) 8t], 

where S is the cross section of the tube. Hence show that D (x) v (x) = const, is the 
condition that the flow of the fluid shall not change the density at any point. 

23. Consider the curve y = f(x) and three equally spaced ordinates at x = a — 5, 
x — a, x = a + 5. Inscribe a trapezoid by joining the ends of the ordinates at 
x = a ± 5 and circumscribe a trapezoid by drawing the tangent at the end of the 
ordinate at x = a and producing to meet the other ordinates. Show that 

S = 2 5/(a), S = 2 d[f(a) + |/"(a) + ^/ (iv) (£)] ' 

S, = 2 s[/(a) + |/"(a) + ^/ (iv) &)] 



TAYLOK'S FORMULA; ALLIED TOPICS 77 

are the areas of the circumscribed trapezoid, the curve, the inscribed trapezoid. 
Hence infer that to compute the area under the curve from the inscribed or cir- 
cumscribed trapezoids introduces a relative error of the order 5 2 , but that to com- 
pute from the relation 8 = £ (2 S Q + SJ introduces an error of only the order of 5 4 . 

24. Let the interval from a to b be divided into an even number 2n of equal 
parts 5 and let the 2 n + 1 ordinates y , y 1 , • • •, y- 2 n at the extremities of the inter- 
vals be drawn to the curve y =f(x). Inscribe trapezoids by joining the ends of 
every other ordinate beginning with 2/ , t/ 2 , and going to y 2n . Circumscribe trape- 
zoids by drawing tangents at the ends of every other ordinate y x , y 3 , • • •, y- 2n -\. 
Compute the area under the curve as 

S = C b f(x)dX = t^. [4 (?; + Vz + • • ■ + V-2n-l) 
Ja O 

+ 2 (y + y 2 + • • • + y 2n ] - y Q - y. 2n ] + R 

by using the work of Ex. 23 and infer that the error R is less than (b— a) 5 4 /( iv >(£)/45. 
This method of computation is known as Simpson'' s Rule. It usually gives accu- 
racy sufficient for work to four or even five figures when 5 = 0.1 and b — a = 1 ; for 
/( iv )(x) usually is small. 

25. Compute these integrals by Simpson's Eule. Take 2n = 10 equal intervals. 
Carry numerical work to six figures except where tables must be used to find f(x) : 

J" ^ cLt f 1 (It 

— = log 2 = 0.69315, (B) / — — = tan-i 1 = -tt = 0.78535, 

ix Jo 1 + x 2 4 

(7) f 5 sin xdx = 1.00000, (5) f - log 10 xdx = 

Jo Ji 

r 1 log(l + x)^ = Q f 11 °g( 1 + a; ) (fa . == Q.82247. 

v ' Jo 1 + x 2 Vi ; Jo X 

The answers here given are the true values of the integrals to five places 

26. Show that the quadrant of the ellipse x = a sin 0, y = b cos is 

s = a ( * Vl — e 2 sin 2 d0 = ± mx f V^ (2 — e 2 ) + i e 2 cos 7™ 

Jo Jo 



dw. 



Compute to four figures by Simpson's Rule with six divisions the quadrants of 
the ellipses : 

(a) e = iV3, s = 1.211 a, (/3) e = \ V2, s = 1.351 a. 

27. Expand s in Ex. 26 into a series and discuss the remainder. 

2 L W \2-4/ 3 V2.4.6/ 5 V 2-4...2n /2n-l J 

1 /l • 3- • • (2n 4- lU 2 e 2n + 2 

R n < -( — ) SeeEx.l8,p.60,andPeirce's"Tables,"p.62. 

1-e 2 \2.4...(2n + 2)/ 2n+l 

Estimate the number of terms necessary to compute Ex. 26 (8) with an error not 
greater than 2 in the last place and compare the labor with that of Simpson's Rule. 

28. If the eccentricity of an ellipse is g^, find to five decimals the percentage 
error made in taking 2 iza as the perimeter. Ans. 0.00694% 



78 DIFFERENTIAL CALCULUS 

29. If the catenary y = c cosh (x/c) gives the shape of a wire of length L sus- 
pended between two points at the same level and at a distance I nearly equal to 
X, find the first approximation connecting X, I, and d, where d is the dip of the 
wire at its lowest point below the level of support. 

30. At its middle point the parabolic cable of a suspension bridge 1000 ft. long 
between the supports sags 50 ft. below the level of the ends. Find the length of 
the cable correct to inches. 

40. Some differential geometry. Suppose that between the incre- 
ments of a set of variables all of which depend on a single variable t 
there exists an equation which is true except for infinitesimals of higher 
order than At = dt, then the equation will be exactly true for the differ- 
entials of the variables. Thus if 

fAx + gAy + hAz + IM + • • • + e 1 + e 2 + • • • = 

is an equation of the sort mentioned and if the coefficients are any func- 
tions of the variables and if e v e 2 , ■ ■ ■ are infinitesimals of higher order 
than dt, the limit of 

J At J At At At At At 

dx dy da 

■ f irt +ff d~t + h irt + l = ' 

or fdx + gdy + hdz -{-ldt = 0; 

and the statement is proved. This result is very useful in writing 
down various differential formulas of geometry where the approximate 
relation between the increments is obvious and where the true relation 
between the differentials can therefore be found. 

For instance in the case of the differential of arc in rectangular coor- 
dinates, if the increment of arc is known to differ from its chord by an 
infinitesimal of higher order, the Pythagorean theorem shows that the 
equation As 2 = Ax 2 + At/ or As 2 = Ax 2 + Af + Az 2 (7) 

is true except for infinitesimals of higher order; and hence 

ds 2 = dx 2 + df or ds 2 = dx 2 + dy 2 + dz 2 . (7') 

In the case of plane polar coordinates, the triangle PP'N (see Fig.) 



^% 




P' 

P^Jm 



has two curvilinear sides PP' and PN and is right- 
angled at N. The Pythagorean theorem may be 
applied to a curvilinear triangle, or the triangle may 
be replaced by the rectilinear triangle PP'N with !Zx 

the angle at N no longer a right angle but nearly so. In either way of 
looking at the figure, it is easily seen that the equation As 2 = A;- 2 -j- r 2 A<f> 2 



TAYLOR'S FORMULA: ALLIED TOPICS 



T9 



which the figure suggests differs from a true equation by an infinitesi- 
mal of higher order; and hence the inference that in polar coordinates 
ds 2 = dr 2 + r*d>4>\ 

The two most used systems of coordinates 
other than rectangular in space are the polar 
or spherical and the cylindrical. In the first 
the distance r = OP from the pole or center, 
the longitude or meridional angle <£, and the 
colatitude or polar angle are chosen as coor- 
dinates ; in the second, ordinary polar coordinates r = OM and <£ in 
the ^?/-plane are combined with the ordinary rectangular z for distance 
from that plane. The formulas of transformation are 

z = r cos 0, 



z 


p , 


x,y,z 

r,<p,d 






rl 


r,cp,z 






y 


z 







i 


y 


Y 


^^5, 


!Xr 






X ^ <P \ 


M 




^ 







■^x 2 + f + z 



„2 



COS" 



■^Jx 2 + if + z 2) 



y = r sin sin <f>, 

x = r sin 6 cos <f>, 

for polar coordinates, and for cylindrical coordinates they are 
z = z, y = r sin <p, 



i V 
<h = tan -1 - 1 

x 



r sm <f>, 



VoJ 2 + 



r 



<f> — tan 



(8) 



(9) 



Formulas such as that 
for the differential of 
arc may be obtained for 
these new coordinates by 
mere transformation of 
(7') according to the rules 
for change of variable. 

In both these cases, 
however, the value of 
ds may be found readily 
by direct inspection of 
the figure. The small 
parallelepiped (figure 
for polar case) of which 
As is the diagonal has 
some of its edges and 
faces curved instead of 
straight; all the angles, 
however, are right angles, 
and as the edges are infinitesimal, the equations certainly suggested as 
holding except for infinitesimals of higher order are 




80 DIFFERENTIAL CALCULUS 

As 2 = Ar 2 + r 2 sin 2 0A<£ 2 + r 2 A0 2 and As 2 = Ar 2 + r 2 ^ 2 + As 2 (10) 
or ds 2 = d72 + 7*sm 2 Odf + r i d0 i and ds 2 = dr 1 + 7*d<t? + dz\ (10') 

To make the proof complete, it would be necessary to show that noth- 
ing but infinitesimals of higher order have been neglected and it might 
actually be easier to transform V dx 2 -\- dy 2 -\- dz 2 rather than give a 
rigorous demonstration of this fact. Indeed the infinitesimal method is 
seldom used rigorously ; its great use is to make the facts so clear to the 
rapid worker that he is willing to take the evidence and omit the proof. 
In the plane for rectangular coordinates with rulings parallel to the 
?/-axis and for polar coordinates with rulings issuing from the pole the 
increments of area differ from 

dA = ydx and dA = ± r 2 d<f> (11) 

respectively by infinitesimals of higher order, and 

= C \jdx and A = _J f ' vhl$ (11') 

«-/X t/0 O 

are therefore the formulas for the area under a curve and between two 
ordinates, and for the area between the curve and two radii. If the plane 
is ruled by lines parallel to both axes or by lines issuing from the pole 
and by circles concentric with the pole, as is customary for double inte- 
gration (§§ 131, 134), the increments of area differ respectively by 
infinitesimals of higher order from 

dA = dxdy and dA = rdrdcft, (12) 

and the formulas for the area in the two cases are 



A 



A = lim Y A.4 = CfdA = I (dxdy, 
A = lim^A.4 = CfdA = jfrdrd^, 



(12') 



where the double integrals are extended over the area desired. 

The elements of volume which are required for triple integration 
(§§ 133, 134) over a volume in space may readily be written down for 
the three cases of rectangular, polar, and cylindrical coordinates. In the 
first case space is supposed to be divided up by planes x = a, y = b, 
z = c perpendicular to the axes and spaced at infinitesimal intervals ; in 
the second case the division is made by the spheres r = a concentric 
with the pole, the planes cf> = b through the polar axis, and the cones 
6 — c of revolution about the polar axis ; in the third case by the cylin- 
ders r = a, the planes <f> = b, and the planes z = c. The infinitesimal 



TAYLOR'S FORMULA; ALLIED TOPICS 81 

volumes into which space is divided then differ from 

dv = dxdydzj dv = r 3 sin ddrd<f>d$, dv = rdrdfydz (13) 

respectively by infinitesimals of higher order, and 



fiidxdydz, J I {>* sin Odrd^dO, j j irdrdfr 



Iz (13') 




are the formulas for the volumes. 

41. The direction of a line in space is represented by the three angles 
which the line makes with the positive directions of the axes or by the 
cosines of those angles, the direction cosines of the line. From the defi- 
nition and figure it appears that 

dx _ du dz 

l = cosa = —, m = cosB=-f-. ra = cosy = — (11) 

ds ds ' dx v 7 

are the direction cosines of the tangent to the arc at the point; of the 
tangent and not of the chord for the reason 
that the increments are replaced by the differ- 
entials. Hence it is seen that for the direc- 
tion cosines of the tangent the proportion 

I : m : n = dx : dy : dz (H') 

holds. The equations of a space curve are 

in terms of a variable parameter t* At the point (x Q} y , z Q ) where 
t = t the equations of the tangent lines would then be 

x — x _ y — y __ z — z Q x — x __ y — y ___ z — z 

(dx\ (dy\ (dz\ ' f(t ) g%) K%)' 

As the cosine of the angle 9 between the two directions given by the 
direction cosines I, ?n, n and V, m', n l is 

cos0 = tl' + mm' + nn', so IV -f- mm' + run) = (16) 

is the condition for the perpendicularity of the lines. Xow if (x, y, z) 
lies in the plane normal to the curve at a? , y , z Q , the lines determined 
by the ratios x : — x Q : y — y Q : z — z Q and (dx) : (dy) : (dz) Q will be per- 
pendicular. Hence the equation of the normal plane is 

(* - ^W, + (y - y a Wy\ + (* - z,){dz\ = o 

/'(y(*-* ) + /(OG/-2/ ) + /<'(y(z-*„) = 0. (17) 

* For the sake of generality the parametric form in t is assumed ; in a particular case a 
simplification might he made hy taking one of the variables as t and one of the functions 
/', g\ h' would then be 1. Thus in Ex. 8 (e), y should be taken as t. 



82 



DIFFERENTIAL CALCULUS 



The tangent plane to the curve is not determinate ; any plane through 
the tangent line will be tangent to the curve. If A be a parameter, the 
pencil of tangent planes is 






A 



y -y* 



-(!+*) 



= 0. 



There is one particular tangent plane, called £Ae osculating plane,which 
is of especial importance. Let 



*0 = /'('«) T + */"(0 ^ + if '"(f) **> ? = * 



<*<*, 



with similar expansions for y and «, be the Taylor developments of 
a?, y, «■ about the point of tangency. When these are substituted in the 
equation of the plane, the result is 



1 ,,[•/"& 

2 L/'C, 



) j v g"W 



(i+*) 



*'(O.J 



6 T 



./'Co) + V('o) (+X) /' 



1Q1 



This expression is of course proportional to the distance from any point 
x, y, z of "the curve to the tangent plane and is seen to be in general of 
the second order with respect to r or ds. It is, however, possible to 
choose for A that value which makes the first bracket vanish. The tan- 
gent plane thus selected has the property that the distance of the curve 
from it in the neighborhood of the point of tangency is of the third order 
and is called the osculating plane. The substitution of the value of A gives 



= (18) 



* - * y-% 


*-*0 




x x 


y-y a 


»-■*„ 


/'(*„) </(h) 


*w 


= or 


W)i 


Wo 


(<k)o 


/"(*„) ff"(Q 


*"(& 




(**). 


(A)„ 


(*). 



01 



(dyd 2 z — dzd 2 y) (x — cc ) + (dzd 2 x — dxd 2 z) (y — y ) 

as the equation of the osculating plane. In case/"(£ )=^"(£ ) = &"(£ ) = 0, 
this equation of the osculating plane vanishes identically and it is neces- 
sary to push the development further (Ex. 11). 

42. For the case of plane curves the curvature is denned as the rate 
at which the tangent turns compared with the description of arc, that 
is, as dcfi/ds if d<j> denotes the differential of the angle through which 
the tangent turns when the point of tangency advances along the curve 
by ds. The radius of curvature R is the reciprocal of the curvature, 
that is, it is ds/d<j>. Then 



deb = d tan* 1 rr > 
ax 



d$ 
ds 



d$dx 
dx ds 



[i+y'J 



, = Ci±A\ (19) 



TAYLOR'S FORMULA; ALLIED TOPICS 83 

"where accents denote differentiation with respect to x. For space curves 
the same definitions are given. If I, m, n and I + dl, m + dm , n -f- dn 
are the direction cosines of two successive tangents, 

cos d<f> = 1(1 + dl) -f- m (m -f- dm) -f n (n + dn). 

But I 2 + »i 2 + n 2 = 1 and (I + dl) 2 + (m + rfm) 2 + (n + ^) 2 = 1. 

Hence tf/ 2 + ^* 2 + «*» a = 2 — 2 cos <£ = (2 sin * <f>) 2 , 



= r rf(2smt») -|' = ^ + fj* + ^ = p , + Bl „ + „* (19 , } 
ds ds 



1_ _ ( C H\ 2 - [~ ^(2sini-<fr) ] 2 _ r// 2 + f/>^ 2 + (fa 2 



where accents denote differentiation with respect to s. 

The torsion of a space curve is denned as the rate of turning of the 
osculating plane compared with the increase of arc (that is, d\f//ds, where 
dip is the differential angle the normal to the osculating plane turns 
through), and may clearly be calculated by the same formula as the 
curvature provided the direction cosines L, M, JS T of the normal to the 
plane take the places of the direction cosines I, m, n of the tangent line. 
Hence the torsion is 

and the radius of torsion R is defined as the reciprocal of the torsion, 
where from the equation of the osculating plane 

L M N 



dyd 2 z — dzd 2 y dzd 2 x — dxoV'z dxd 2 y — dyd 2 x 

= , 1 (20') 

V sum of squares 

The actual computation of these quantities is somewhat tedious. 

The vectorial discussion of curvature and torsion (§ 77) gives a better insight 
into the principal directions connected with a space curve. These are the direction 
of the tangent, that of the normal in the osculating plane and directed towards 
the concave side of the curve and called the principal normal, and that of the 
normal to the osculating plane drawn upon that side which makes the three direc- 
tions form a right-handed system and called the binormal. In the notations there 
given, combined with those above, 

r = xi + yi + zk, t = Zi + m] + nk, c = Xi + /*j + vk, n = Li + M] + Nk, 

where X, ix, v are taken as the direction cosines of the principal normal. Now dt 
is parallel to c and dn is parallel to — c. Hence the results 

Jo 

(21) 



dl dm dn ds 




dL dlT dN ds 


— rr ~ — — — 


and 


— — zz: — 


A M 9 R 




X A* " R 



above formulas, — = 
right-handed 


x' y' z' 
x" y" z" 
x"' y'" z'" 


i usual formulas, — = — 
left-handed R 


x y' z' 
x" y" z" 
x"' y'" z'" 


x" 2 + y" 2 + z" 2 


x "-2 _j_ y"2 _|_ z "2 



84 DIFFEKENTIAL CALCULUS 

follow from dc/ds = C and dn/ds = T. Now dc is perpendicular to c and hence in 
the plane of t and n; it may be written as dc = (t»dc)t+ (n.dc)n. But as t.c = n.c = 0, 
t.dc = — cdt and n.cZc = — ccfa. Hence 

dc=- (cdt)t - (cdn)n = - Ctds + Tnds = -lds + -ds. 

E R 

d\ I L djx m , M dp n N /ft „ 

HellCe di=-fi + R' ds = -R + -R' Is=-R + R- <22> 

Formulas (22) are known as FreneVs Formulas; they are usually written with — R 

in the place of R because a left-handed system of axes is used and the torsion, being 

an odd function, changes its sign when all the axes are reversed. If accents denote 

differentiation by s, 



(23) 



EXERCISES 

1. Show that in polar coordinates in the plane, the tangent of the inclination 
of the curve to the radius vector is rdcp/dr. 

2. Verify (10), (10') by direct transformation of coordinates. 

3. Fill in the steps omitted in the text in regard to the proof of (10), (10') by 
the method of infinitesimal analysis. 

4. A rhumb line on a sphere is a line which cuts all the meridians at a constant 
angle, say a. Show that for a rhumb line sin &dcj> = tan add and ds = r sin adO. 
Hence find the equation of the line, show that it coils indefinitely around the 
poles of the sphere, and that its total length is irr sec a. 

5. Show that the surfaces represented by F($, 6) = and F(r, 6) = in polar 
coordinates in space are respectively cones and surfaces of revolution about the 
polar axis. What sort of surface would the equation F(r, $) = represent ? 

6. Show accurately that the expression given for the differential of area in 
polar coordinates in the plane and for the differentials of volume in polar and 
cylindrical coordinates in space differ from the corresponding increments by in- 
finitesimals of higher order. 

7. Show that — , r — , rsin# — are the direction cosines of the tangent to a 

ds ds ds 

space curve relative to the radius, meridian, and parallel of latitude. 

8. Find the tangent line and normal plane of these curves. 

(a) xyz = 1, y 2 = x at (1, 1, 1), (fi) x = cos t, y = sin£, z = kt, 

(y) 2 ay = x 2 , 6 a 2 z = x 3 , (5) x = t cos t, y — t sin t, z = kt, 

(e) y = x 2 , z 2 = 1 - y, (f) x 2 + y 2 + z 2 = a 2 , x 2 + y 2 + 2 ax = 0. 

9. Find the equation of the osculating plane in the examples of Ex. 8. Note 
that if x is the independent variable, the equation of the plane is 

(dy d 2 z dz d 2 y\ . . (d 2 z\ . . , (d 2 y\ . . . 



TAYLOR'S FORMULA; ALLIED TOPICS 85 

10. A space curve passes through the origin, is tangent to the x-axis, and has 
z — as its osculating plane at the origin. Show that 

x = tf'(0) + \ t 2 £"(0) + • • • , y = l *V'(0) + . . . , z = i t&h"'(0) + • • . 

will be the form of its Maclaurin development if t = gives x = y = z = 0. 

11. If the 2d, 3d. • • • . (n — l)st derivatives of /, gr, 7i vanish for t = f but not 
all the nth derivatives vanish, show that there is a plane from which the curve 
departs by an infinitesimal of the (n + l)st order and with which it therefore 
has contact of order n. Such a plane is called a hyperosculating plane. Find its 
equation. 

12. At what points if any do the curves (/3), (7), (e), (f), Ex. 8 have hyperoscu- 
lating planes and what is the degree of contact in each case ? 



13. Show that the expression for the radius of curvature is 

B ~ [r 2 + g>-l + lx ' 2 f 

where in the first case accents denote differentiation by s, in the second by t. 



14. Show that the radius of curvature of a space curve is the radius of curva- 
ture of its projection on the osculating plane at the point in question. 

15. From Frenet's Formulas show that the successive derivatives of x are 

, , „ „ X ,/, * *#' I , & L 

x = L x = I = — , x = = X 1 , 

P BE 2 E 2 P 2 BR 

where accents denote differentiation by s. Show that the results for y and z are 
the same except that m. n, 31 or n, v, X take the places of I, X, L. Hence infer 
that for the nth derivatives the results are 

BOO = W 1 + \P 2 + iP 3? y(n) - m p i + M p 2 + MP 3 , 2(») = nP 1 + vP, + NP 3 , 

where P x , P 2 , P 3 are rational functions of P and R and their derivatives by s. 

16. Apply the foregoing to the expansion of Ex. 10 to show that 

1 , s 2 B' , s 3 
j; = s s 3 + • • • , y = s 3 + • • ■ , z == 1- • • • . 

6P 2 ' 21? 6B 2 ' 6 PR 

where B and R are the values at the origin where s = 0. I = /j. = X = 1. and the 
other six direction cosines m, n, X, v, X, Ji" vanish. Find s and write the expan- 
sion of the curve of Ex. 8 (7) in this form. 

17. Xote that the distance of a point on the curve as expanded in Ex. 16 from 
the sphere through the origin and with center at the point (0. P. P'R) is 



'x 2 + {y - By 2 + {z - P'R) 2 - vP 2 + P' 2 R 2 

(x 2 + if- - 2 By + z 2 - 2 P'Rz) 



Vz 2 + (y - B) 2 + (z - P'R) 2 + VP 2 + P' 2 R 2 

and consequently is of the fourth order. The curve therefore has contact of the 
third order with this sphere. Can the equation of this sphere be derived by a 
limiting process like that of Ex. 18 as applied to the osculating plane ? 



86 



DIFFEKENTIAL CALCULUS 



18. The osculating plane may be regarded as the plane passed through three 
consecutive points of the curve ; in fact it is easily shown that 



lim 

Sx, Sy, Sz 
&x. A?/, Sz 
approach 



X 




y 


z 




x 


+ dx 


y + Sy 


z o 
z o 


+ 8z 



x + Ax y + Ay z + Az 



x-x y — y 
(dx) (dy) 
(d*x) (d*y) 



z-z 

(dz) 

(d*z) 



0. 



19. Express the radius of torsion in terms of the derivatives of x, y, z by t 
(Ex. 10, p. 67). 

20. Find the direction, curvature, osculating plane, torsion, and osculating 
sphere (Ex. 17) of the conical helix x = t cos t, y = t sin t, z = kt at t = 2 -k. 

21. Upon a plane diagram which shows As, Ax, Ay, exhibit the lines which 
represent ds, dx, dy under the different hypotheses that x, y, or s is the independ- 
ent variable. 



CHAPTER IV 
PARTIAL DIFFERENTIATION; EXPLICIT FUNCTIONS 

43. Functions of two or more variables. The definitions and theo- 
rems about functions of more than one independent variable are to a 
large extent similar to those given in Chap. II for functions of a single 
variable, and the changes and difficulties which occur are for the most 
part amply illustrated by the case of two variables. The work in the 
text will therefore be confined largely to this case and the generaliza- 
tions to functions involving more than two variables may be left as 
exercises. 

If the value of a variable z is uniquely determined when the values 
(x, y) of two variables are known, z is said to be a function z = f(x, y) 
of the two variables. The set of values [_(x, y)~\ or of points P(x, y) of 
the ary-plane for which z is defined may be any set, but usually consists 
of. all the points in a certain area or region of the plane bounded by 
a curve which may or may not belong to the region, just as the end 
point s of an inte rval may or may not belong to it. Thus the function 
1/ Vl — x 2 — if is defined for all points within the circle x 2 + y 2 = 1, 
but not for points on the perimeter of the circle. For most purposes it 
is sufficient to think* of the boundary of the region of definition as a 
polygon whose sides are straight lines or such curves as the geometric 
intuition naturally suggests. 

The first way of representing the function z =f(x, y) geometrically 
is by the surface z =f(x, y), just as y = f(x) was represented by a curve. 
This method is not available for u =f(x, y, z), a function of three vari- 
ables, or for functions of a greater number of variables ; for space has 
only three dimensions. A second method of representing the function 
z=f(x, y) is by its contour lines in the a-y-plane, that is, the curves 
f(x, y) = const, are plotted and to each curve is attached the value of 
the constant. This is the method employed on maps in marking heights 
above sea level or depths of the ocean below sea level. It is evident that 
these contour lines are nothing but the projections on the ay-plane 
of the curves in which the surface z = f(xf y) is cut by the planes 
z = const. This method is applicable to functions u —fix, y, z) of 
three variables. The contour surfaces u = const, which are thus obtained 

87 



'88 



DIFFEBENTIAL CALCULUS 



are frequently called equipotential surfaces. If the function is single 
Talued, the contour lines or surfaces cannot intersect one another. 

The function z = f(x, y) is continuous for (a, b) when either of the 
following equivalent conditions is satisfied : 

1°. limf(x, y) = f(a, b) or lim/(a;, y) = /(lim x, lim y), 
no matter hoiv the variable point P(x, y) approaches (a., b). 

2°. If for any assigned e, a number 8 may be found so that 
\f(x, y) — f(a, b) | < e when | x — a [ < 8, | y — b \ < 8. 
Geometrically this means that if a square with (a, b) as center and 
with sides of length 2 8 parallel to the axes be drawn, 
the portion of the surface z = f(x, y) above the 
square will lie between the two planes z=f(a, b) ± e. 
Or if contour lines are used, no line f(x, y) = const, 
where the constant differs from f(a, b) by so much 
as e will cut into the square. It is clear that in place 
of a square surrounding (a, b) a circle of radius 8 or any other figure 
which lay within the square might be used. 

44. Continuity examined. From the definition of continuity just given and 
from the corresponding definition in § 24, it follows that if /(x, y) is a continuous 
function of x and y for (a, 6), then/(x, b) is a continuous function of x for x = a 
and /(a, y) is a continuous function of y for y — b. That is, if / is continuous in 
x and y jointly, it is continuous in x and y severally. It might be thought that 
conversely if f(x, b) is continuous for x = a and /(a, y) for y = b, f(x, y) would 
be continuous in (x, y) for (a, b) . That is, if / is continuous in x and y severally, 
it would be continuous in x and y 

"5 



Y 




/(<x,b)+e 




25 j 


\\ 


s 




\ 


f(a,b 


/'(a, 6) 
-e 







2d 


X 



jointly. A simple example will show 
that this is not necessarily true. Con- 
sider the case 

z=f(%, y) 



x 2 + y 2 




x + y 
/(0, 0) = o 

and examine z for continuity at 
(0, 0). The functions f(x, 0) = x, 
and/(0, y) =y are surely continuous 
in their respective variables. But the surface z =f(x, y) is a conical surface (except 
for the points of the 2-axis other than the origin) and it is clear that P (x, y) may 
approach the origin in such a manner that z shall approach any desired value. 
Moreover, a glance at the contour lines shows that they all enter any circle or 
square, no matter how small, concentric with the origin. If P approaches the origin 
along one of these lines, z remains constant and its limiting value is that constant. 
In fact by approaching the origin along a set of points which jump from one con- 
tour line to another, a method of approach may be found such that z approaches 
no limit whatsoever but oscillates between wide limits or becomes infinite. Clearly 
the conditions of continuity are not at all fulfilled by z at (0, 0). 



PAETIAL DIFFERENTIATION": EXPLICIT 



Double limits. There often arise for consideration expressions like 



lim I" lim /(x, y)~\, lim [" lim f(x, y)l, 



(1) 



where the limits exist whether x first approaches its limit, and then y its limit, or 
vice versa, and where the question arises as to whether the two limits thus obtained 
are equal, that is, whether the order of taking the limits in the double limit may 
be interchanged. It is clear that if the function /(x, y) is continuous at (a, 5), the 
limits approached by the two expressions will be equal ; for the limit of /(x, y) is 
/(a, b) no matter how (x, y) approaches (a, b). If / is discontinuous at (a, 6), it 
may still happen that the order of the limits in the double limit may be inter- 
changed, as was true in the case above where the value in either order was zero ; 
but this cannot be affirmed in general', and special considerations must be applied 
to each case when /is discontinuous. 

Varieties of regions * For both pure mathematics and physics the classification 
of regions according to their connectivity is important. Consider a finite region R 
bounded by a curve which nowhere cuts itself. (For the present 
purposes it is not necessary to enter upon the subtleties of the 
meaning of "curve" (see §§127-128); ordinary intuition will 
suffice.) It is clear that if any closed curve drawn in this region 
had an unlimited tendency to contract, it could draw together 
to a point and disappear. On the other hand, if R' be a region 
like R except that a portion has been removed so that R' is 
bounded by two curves one within the other, it is clear that 
some closed curves, namely those which did not encircle the 
portion removed, could shrink away to a point, whereas other 
closed curves, namely those which encircled that portion, could 
at most shrink down into coincidence with the boundary of that 
portion. Again, if two portions are removed so as to give rise 
to the region R", there are circuits around each of the portions 
which at most can only shrink down to the boundaries of those 
portions and circuits around both portions which can shrink down to the bounda- 
ries and a line joining them. A region like E, where any closed curve or circuit 
may be shrunk away to nothing is called a simply connected region ; whereas regions 
in which there are circuits which cannot be shrunk away to nothing are called 
multiply connected regions. 

A multiply connected region may be made simply connected by a simple device 
and convention. For suppose that in R' a line were drawn connecting the two 
bounding curves and it were agreed that no curve or circuit drawn within R' should 
cross this line. Then the entire region would be surrounded by a 
single boundary, part of which would be counted twice. The figure 
indicates the situation. In like manner if two lines were drawn in 
R" connecting both interior boundaries to the exterior or connecting 
the two interior boundaries together and either of them to the outer 
boundary, the region would be rendered simply connected. The entire region 
would have a single boundary of which parts would be counted twice, and any 
circuit which did not cross the lines could be shrunk away to nothing. The lines 





* The discussion from this point to the end of 
123-126. 



45 may be connected with that of 



90 



DIFFERENTIAL CALCULUS 




thus drawn in the region to make it simply connected are called cuts. There is no 
need that the region be finite ; it might extend off indefinitely in some directions 
like the region between two parallel lines or between the sides of an angle, or like 
the entire half of the x?/-plane for which y is positive. In such cases the cuts may 
be drawn either to the boundary or off indefinitely in such a way as not to meet 
the boundary. 

45. Multiple valued functions. If more than one value of z corresponds to the 
pair of values (x, y), the function z is multiple valued, and there are some note- 
worthy differences between multiple valued functions of one variable and of several 
variables. It was stated (§ 23) that multiple 
valued functions were divided into branches 
each of which was single valued. There are 
two cases to consider when there is one vari- 
able, and they are illustrated in the figure. 
Either there is no value of x in the interval 
for which the different values of the function 
are equal and there is consequently a number 
_D which gives the least value of the difference 
between any two branches, or there is a value of x for which different branches 
have the same value. Now in the first case, if x changes its value continuously and 
if fix) be constrained also to change continuously, there is no possibility of passing 
from one branch of the function to another ; but in the second case such change is 
possible for, when x passes through the value for which the branches have the same 
value, the function while constrained to change its value continuously may turn off 
onto the other branch, although it need not do so. 

In the case of a function z =f(x, y) of two variables, it is not true that if the 
values of the function nowhere become equal in or on the boundary of the region 
over which the function is defined, then it is impossible to pass continuously from 
one branch to another, and if P (x, y) describes any fc 

continuous closed curve or circuit in the region, the 
value of /(x, y) changing continuously must return to 
its original value when P has completed the descrip- 
tion of the circuit. For suppose the function z be a 
helicoidal surface z = a -tan- 1 (?//x), or rather the por- 
tion of that surface between two cylindrical surfaces 
concentric with the axis of the helicoid, as is the case 
of the surface of the screw of a jack, and the circuit 
be taken around the inner cylinder. The multiple num- 
bering of the contour lines indicates the fact that the 
function is multiple valued. Clearly, each time that 
the circuit is described, the value of z is increased by the amount between the suc- 
cessive branches or leaves of the surface (or decreased by that amount if the circuit 
is described in the opposite direction). The region here dealt with is not simply 
connected and the circuit cannot be shrunk to nothing — which is the key to the 
situation. 

Theorem. If the difference between the different values of a continuous mul- 
tiple valued function is never less than a finite number D for any set (x, y) of 
values of the variables whether in or upon the boundary of the region of defini- 
tion, then the value /(x, y) of die function, constrained to change continuously, 





PAETIAL DIFFERENTIATION; EXPLICIT 91 

will return to its initial value when the point P(x, ?/), describing a closed curve 
which can be shrunk to nothing, completes the circuit and returns to its starting 
point. 

Xow owing to the continuity of / throughout the region, it is possible to find a 
number 5 so that | f(x, y) — f(x', y') \ < e when | x — x' \ < 5 and | y — y' \ < 5 no matter 
what points of the region (x, y) and (x', y') may be. Hence the values of / at any 
two points of a small region which lies within any circle of radius \ 5 cannot differ 
by so much as the amount D. If, then, the circuit is so small 
that it may be inclosed within such a circle, there is no possi- 
bility of passing from one value of / to another when the circuit 
is described and / must return to its initial value. Next let 
there be given any circuit such that the value of / starting from 
a given value /(x, y) returns to that value when the circuit has 
been completely described. Suppose that a modification were 
introduced in the circuit by enlarging or diminishing the inclosed area by a small 
area lying wholly within a circle of radius \ 8. Consider the circuit ABC IDE A and 
the modified circuit ABCDEA. As these circuits coincide except for the arcs BCD 
and BCD, it is only necessary to show that/ takes on the same value at D whether 
D is reached from B by the way of C or by the way of C . But this is necessarily 
so for the reason that both arcs are within a circle of radius \ 8. 
Then the value of / must still return to its initial value /(x, y) 
when the modified circuit is described. Now to complete the 
proof of the theorem, it suffices to note that any circuit which 
can be shrunk to nothing can be made up by piecing together a 
number of small circuits as shown in the figure. Then as the 
change in /around any one of the small circuits is zero, the change must be zero 
around 2, 3, 4, • • • adjacent circuits, and thus finally around the complete large 
circuit. 

Reducibility of circuits. If a circuit can be shrunk away to nothing, it is said to 
be reducible ; if it cannot, it is said to be irreducible. In a simply connected region 
all circuits are reducible ; in a multiply connected region there are an infinity of 
irreducible circuits. Two circuits are said to be equivalent or reducible to each 
other when either can be expanded or shrunk into the other. The change in the 
value of / on passing around two equivalent circuits from A to A 
is the same, provided the circuits are described in the same direc- 
tion. Tor consider the figure and the equivalent circuits AC A 
and AC A described as indicated by the large arrows. It is clear 
that either may be modified little by little, as indicated in the 
proof above, until it has been changed into the other. Hence the C 

change in the value of / around the two circuits is the same. Or, as another proof, 
it may be observed that the combined circuit AC AC' A, where the second is 
described as indicated by the small arrows, may be regarded as a reducible circuit 
which touches itself at A . Then the change of / around the circuit is zero and / 
must lose as much on passing from A to A by C as it gains in passing from A to 
A by C. Hence on passing from A to A by C in the direction of the large arrows 
the gain in /must be the same as on passing by C. 

It is now possible to see that any circuit ABC may be reduced to circuits around 
the portions cut out of the region combined with lines going to and from A and the 
boundaries. The figure shows this ; for the circuit ABCBADC'DA is clearly 





DIFFERENTIAL CALCULUS 




reducible to the circuit AC A. It must not be forgotten that although the lines AB 
and BA coincide, the values of the function are not necessarily the same on AB 
as on BA but differ by the amount of change introduced in 
/ on passing around the irreducible circuit BC'B. One of the 
cases which arises most frequently in practice is that in 
which the successive branches of fix, y) differ by a constant 
amount as in the case z = tan- 1 (y/x) where 2 tt is the differ- 
ence between successive values of z for the same values of the 

variables. If now a circuit such as A BC'B A be considered, where it is imagined 
that the origin lies within BC'B, it is clear that the values of z along AB and 
along BA differ by 2 ir, and whatever z gains on passing from A to 
B will be lost on passing from B to A, although the values through 
which z changes will be different in the two cases by the amount 
2 7r. Hence the circuit ABC'BA gives the same changes for z as 
the simpler circuit BC'B. In other words the result is obtained 
that if the different values of a multiple valued function for the same 
values of the variables differ by a constant independent of the values of 
the variables, any circuit may be reduced to circuits about the bound- 
aries of the portions removed ; in this case the lines going from the point A to the 
boundaries and back may be discarded. 




EXERCISES 

1. Draw the contour lines and sketch the surfaces corresponding to 
x + V „/a A^ a /m „ vy 



(a) z = 



2(0,0) = 0, 



(/3) 



z(0, 0) = 0. 



x — y • '■ ' ' v ' x + y 

Note that here and in the text only one of the contour lines passes through the 
origin although an infinite number have it as a frontier point between two parts 
of the same contour line. Discuss the double limits lim Km z, lim lim z. 

2. Draw the contour lines and sketch the surfaces corresponding to 



(a) z = 



x 2 + y< 



(0) 2 



(7) Z 



x 2 + 2y*-l 



2y x 2x 2 + y 2 — 1 

Examine particularly the behavior of the function in the neighborhood of the 
apparent points of intersection of different contour lines. Why apparent ? 

3. State and prove for functions of two independent variables the generaliza- 
tions of Theorems 6-11 of Chap. II. Note that the theorem on uniformity is proved 
for two variables by the application of Ex. 9, p. 40, in almost the identical manner 
as for the case of one variable. 

4. Outline definitions and theorems for functions of three variables. In partic- 
ular indicate the contour surfaces of the functions 



(a) u = 



x + y + 2z 



(18) u = 



x 2 + y 2 + z 2 



(y) u 



xy 

z 



x — y — z v ' x + y + z 

and discuss the triple limits as x, y, z in different orders approach the origin. 

5. Let z — P(x, y)/Q(x, y), where P and Q are polynomials, be a rational func- 
tion of x and y. Show that if the curves P = and Q = intersect in any points, 
all the contour lines of z will converge toward these points ; and conversely show 



PARTIAL DIFFERENTIATION; EXPLICIT 93 

that if two different contour lines of z apparently cut in some point, all the contour 
lines will converge toward that point, P and Q will there vanish, and z will be 
undefined. 

6. If D is the minimum difference between different values of a multiple valued 
function, as in the text, and if the function returns to its initial value plus D' j== Z) 
when P describes a circuit, show that it will return to its initial value plus D'^D 
when P describes the new circuit formed by piecing on to the given circuit a small 
region which lies within a circle of radius \ 5. 

7. Study the function z = tan-- 1 (y/»), noting especially the relation between 
contour lines and the surface. To eliminate the origin at which the function is not 
defined draw a small circle about the point (0, 0) and observe that the region of 
the whole xy-plane outside this circle is not simply connected but may be made so 
by drawing a cut from the circumference off to an infinite distance. Study the 
variation of the function as P describes various circuits. 

8. Study the contour lines and the surfaces due to the functions 

1 — x 2 
(a) z = tan- x xy, (j3) z = tan- 1 , (7) z = sin- 1 (x — y) . 

i-y- 

Cut out the points where the functions are not defined and follow the changes in 
the functions about such circuits as indicated in the figures of the text. How may 
the region of definition be made simply connected ? 

9. Consider the function z — tan- 1 (P/Q) where P and Q are polynomials and 
where the curves P = and Q = intersect in n points (a 19 b x ), (a 2 , 6 2 ), • • • , (a n , b n ) 
but are not tangent (the polynomials have common solutions which are not mul- 
tiple roots). Show that the value of the function will change by 2Tar if (x, y) 
describes a circuit which includes k of the points. Illustrate by taking for P/Q 
the fractions in Ex. 2. 

10. Consider regions or volumes in space. Show that there are regions in which 
some circuits cannot be shrunk away to nothing ; also regions in which all circuits 
may be shrunk away but not all closed surfaces. 

46. First partial derivatives. Let z=f(x,y) be a single valued 
function, or one branch of a multiple valued function", denned for (a, b) 
and for all points in the neighborhood. If y be given the value b, 
then z becomes a function f(x, b) of x alone, and if that function has a 
derivative for x = a, that derivative is called the partial derivative of 
z =f(x, y) with respect to x at (a, b). Similarly, if x is held fast and 
equal to a and if f(a, y) has a derivative when y = b, that derivative is 
called the partial derivative of z with respect to y at (a, b). To obtain 
these derivatives formally in the case of a given function f(x, y) it is 
merely necessary to differentiate the function by the ordinary rules, 
treating yas a constant when finding the derivative with respect to x 
and x as a constant for the derivative with respect to y. Notations are 



94 DIFFERENTIAL CALCULUS 

for the ^-derivative with similar ones for the ^/-derivative. The partial 
derivatives are the limits of the quotients 

lim f(a + k, b) -f ( a, f >) ! lim /( ff , S + /,■)-/(„,*) , (2) 

provided those limits exist. The application of the Theorem of the 
Mean to the functions f(x, b) and f(a, y) gives 

f(a + h, b) - f(a, b) = hf x (a + Ofi, b), 0<6 1 <1, 
f(a, b ■ + *) - f(a, b) = kf y (a, b + 2 k), 0< 2 < 1, 

under the proper but evident restrictions (se'e § 26). 

Two comments may be made. First, some writers denote the partial derivatives 
by the same symbols dz/dx and dz/dy as if z were a function of only one variable 
and were differentiated with respect to that variable ; and if they desire especially 
to call attention to the other variables which are held constant, they affix them as 
subscripts as shown in the last symbol given (p. 93). This notation is particularly 
prevalent in thermodynamics. As a matter of fact, it would probably be impos- 
sible to devise a simple notation for partial derivatives which should be satisfac- 
tory for all purposes. The only safe rule to adopt is to use a notation which is 
sufficiently explicit for the purposes in hand, and at all times to pay careful atten- 
tion to what the derivative actually means in each case. Second, it should be noted 
that for points on the boundary of the region of definition of f(x, y) there may be 
merely right-hand or left-hand partial derivatives or perhaps none at all. For it 
is necessary that the lines y = b and x — a cut into the region on one side or the 
other in the neighborhood of (a, b) if there is to be a derivative even one-sided ; 
and at a corner of the boundary it may happen that neither of these lines cuts 
into the region. 

Theorem. If f(x, y) and its derivatives f' x and f' y are continuous func- 
tions of (x, y) in the neighborhood of (a, b), the increment Af may be 
written in any of the three forms 

*f = f(a + h,b + ky-f(d,_b) 

= hf x (a + eji, b) + k/; (a 4- h, b + e 2 k) 

= hf x (a + Oh, b + Ok) + kf' y (a 4- Oh, b + Ok) 
= hf x (c h b) 4- kf y (a, b) 4- Qi 4- y, 
where the 0's are proper fractions, the £'s infinitesimals. 
To prove the first form, add and subtract /(a + h, b) ; then 

Af=[f(a + h, b)-f(a, b)] + [/(a + h, b + k) -f(a + h, 6)] 

= /*/; (a + eji, b) + kfy (a + h,b + e 2 k) 

by the application of the Theorem of the Mean for functions of a single variable 
(§§ 7, 26). The application may be made because the function is continuous and 
the indicated derivatives exist. Now if the derivatives are also continuous, they 
may be expressed as 

/; (a + <y>, b) = /; (a, b) + &, /; (a + h, b + e 2 k) = /; (a, b) + 1 2 






PARTIAL DIFFERENTIATION; EXPLICIT 95 

where £ v f 2 may be made as small as desired by taking h and k sufficiently small. 
Hence the third form follows from the first. The second form, which is symmetric 
in the increments h, k, may be obtained by writing x = a + th and y = b + tk. 
Then/(x, y) = <i> (£). As/ is continuous in (x, ?/), the function $ is continuous in t 
and its increment is 



A$ = f(a + t + Ath, b + t + Atk) —f(a + th, b + tk). 

This may be regarded as the increment of / taken from the point (x, y) with At • h 
and At • k as increments in x and y. Hence A<i> may be written as 

A$ = At • hf x (a + tfi, b + £&) + A. • Jcfy (a + th, b + tk) + ^At ■ /i + f 2 A« • A;. 

Now if A<£ be divided by A£ and,A£ be allowed to approach zero, it is seen that 

A$ , , d$> 
lim — = hf x (a + to, 6 + *&) + fc/ y ' (a + th, b + &) = 

The Theorem of the Mean may now be applied to <£ to give <£ (1) — <£ (0) = 1 • <£'(#), 
and hence 

*(1) - $(0) =f(a + h,b + k)-f(a, b) 

= A/= ft/ x '(a + Oh, b+0k) + kfy{a + 0ft, b + Ok). 

47. The partial differentials of /may be defined as 



rf_ f = f 'Ax, so that <fo; = A;z, 
<"■■"' ' dx dx 

d,f = f^y, so that % = Ay, ^='f£ 



(5) 



where the indices x and y introduced in d x f and d y f indicate that x and 
y respectively are alone allowed to vary in forming the corresponding 
partial differentials. The total differential 

df = dj + dj = 8 £dx + ^dy, (6) 

which is the sum of the partial differentials, may be defined as that 
sum; but it is better defined as that part of the increment 

a/ = c £ a x + % a?j + ^ Ax + i ^ y (T ) 

which is obtained by neglecting the terms £ x Ax + £ 2 Ay, which are of 
higher order than Ax and Ay. The total differential may therefore be 
computed by finding the partial derivatives, multiplying them respec- 
tively by dx and dy, and adding. 

The total differential of z = f(x, y) may be formed for (x , y ) as 



-*•=(!)„<*-* >+&)<*-*>> 



(8) 



where the values x — x and y — y are given to the independent, differ- 
entials dx and dy, and df= dz is written as z — z Q . This, however, is 



96 



DIFFERENTIAL CALCULUS 



the equation of a plane since x and y are independent. The difference 
A/ — df which measures the distance from the plane to the surface 
along a parallel to the 2-axis is of higher order than Va^c 2 + A?/ 2 ; for 



*f-df 



^/Ax 2 + A^ 2 



LAx + J Ay 



<KM£ 2 | = °- 



V Ax 2 + Ay 2 

Hence the plane (8) will be denned as the tangent plane at (x Q , y , z Q ) 
to the surface z = f(x, y). The normal to the plane is 



y-y, 



(9) 



which will be denned as the normal to the surface at (x , y , « ). The 
tangent plane will cut the planes y = y and x = x in lines of which 
the slope is f^ and f . The surface will cut these planes in curves 
which are tangent to the lines. 

In the figure, PQSR is a portion of the 
surface z =f(x, y) and pt'TT" is a cor- 
responding portion of its tangent plane 
at P(x Q , y Q , z Q ). Now the various values 
may be read off. 

P'Q = Kf, 

P'T> = d x f 

P"R = \f, 
P"T" = d y f 

N'S = Af 
N<T=df=d x f+d y f. 

48. If the variables x and y are expressed as x = <j>(t) and y — if/(t) 
so that f(x, y) becomes a function of t, the derivative of / with respect 
to t is found from the expression for the increment of/. 

Af = dfAx of Ay Ax Ay 

At ex At dy At 4l At 4-2 At 

Af df df dx , of dy „„ 

or Inn -J- = -j- = f-— + f--£- (10) 

At = o At dt dx dt dy dt 

The conclusion requires that x and y should have finite derivatives with 
respect to t. The differential of f as a function of t is 

df = % dt = d l^ cU + %^dt = %dx + d £.dy (11) 



PP' = Ax, 
P'T'/PP* =% 

PP" = Ay, 

p"t"/pp"=/;, 

p'T' -4- p"T" = N'T 




dt 



and hence it appears that the differential has the same form as the total 
differential. This result will be generalized later. 




PARTIAL DIFFERENTIATION; EXPLICIT 97 

As a particular case of (10) suppose that x and y are so related that 
the point (x, y) moves along a line inclined at an angle t to the cc-axis. 
If s denote distance along the line, then 

x = x Q + s cos t, y = y -f- s sin r, dx = cos rds, dy = sin rds (12) 
df df dx cf dy . 

and * = £ S + ty ^ =f* C ° S T + ^ Sm T ' < 13 > 

The derivative (13) is called the directional derivative of f in the direc- 
tion of the line. The partial derivatives f' x , f' y are the particular direc- 
tional derivatives along the directions of the a?-axis and ?/-axis. The 
directional derivative of / in any direction is the rate of increase of 
/ along that direction ; if z = f(x, y) be inter- 
preted as a surface, the directional derivative is 
the slope of the curve in which a plane through 
the line (12) and perpendicular to the a*?/-plane 
cuts the surface. If f(x, y) be represented by 
its contour lines, the derivative at a point - 
(x, y) in any direction is the limit of the ratio 
A f/As = A C/As of the increase of /, from one contour line to a neigh- 
boring one, to the distance between the lines in that direction. It is 
therefore evident that the derivative along any contour line is zero and 
that the derivative along the normal to the contour line is greater than 
in any other direction because the element dn of the normal is less than 
ds in any other direction. In fact, apart from infinitesimals of higher 

° rdei '' ^ , Af V , df df ' 

Hence it is seen that the derivative along any direction may be found 
by multiplying the derivative along the normal by the cosine of the angle 
between that direction and the normal. The derivative along the normal 
to a contour line is called the normal derivative of /and is, of course, 
a function of (x, y). 

49. Next suppose that u =f(x, y, z, ■ • •) is a function of any number 
of variables. The reasoning of the foregoing paragraphs may be 
repeated without change except for the additional number of variables. 
The increment of / will take any of the forms 

V = /0 + M + k, c + l,...) -f(a, b,e,~ •) 

= Kf' x {a + BJi, b,c,-.-) + hf y (a + h, b + OJc, c, ■ • •) 

+ /f;(« + M + A-,c + 3 /,---) + --- 

= h£ + kf; + K + • • • + C t h + l 2 k ■+ 1} + • • ., 



98 DIFFERENTIAL CALCULUS 

and the total differential will naturally be defined as 

* #=i*+g*'+f£*+"-. ( 16 ) 

and finally if x, y, z, • • • be functions of t, it follows that 

of = d^dx_^dfdy_^dfdz^ 

dt dx dt dy dt dz dt ^ ' 

and the differential of /as a function of t is still (16). 

If the variables x, y, z, • • ■ were expressed in terms of several new 
variables r, s, • • • , the function / would become a function of those vari- 
ables. To find the partial derivative of / with respect to one of those 
variables, say r, the remaining ones, s, • • • , would be held constant and 
/ would for the moment become a function of r alone, and so would x, 
y, z, •••'. Hence (17) may be applied to obtain the partial derivatives 

dr dx dr dy dr dz dr 

df dfdx dfdy dfdz (18) 

and ^" = ^~^ + ^ _ "5~ + ^ _ "^ + ---5 etc. 

cs ex cs cy ds oz os 

These are the formulas for change of variable analogous to (4) of § 2. 
If these equations be multiplied by Ar, As, ■ • • and added, 

df df df (dx dx \ df (dy 

-f Ar + -f As + . . . = -f — Ar + tt- As + ...) + ^- ( / Ar + • 

Or os cx\or cs J oy\dr 

for when r, s, • • • are the independent variables, the parentheses above 
are dx, dy, dz, • • • and the expression on the left is df. 

Theorem. The expression of the total differential of a function of 
x, y, z, • ■ • as df = f'Jix + f y dy -f- f' z dz -f- • ■ • is the same whether x, y, 
z, • • • are the independent variables or functions of other independent 
variables r, s, • - •; it being assumed that all the derivatives which occur, 
whether of f by x, y, z, • • • or of x, y, z, • • • by r, s, • • • , are continuous 
functions. 

By the same reasoning or by virtue of this theorem the rules 

d (eu) = cdu, d (u -f- v — w) = du + dv — dw, 
d (uv) = udv + vdu, . d(-\= ^ ? 

of the differential calculus will apply to calculate the total differential 
of combinations or functions of several variables. If by this means, or 
any other, there is obtained an expression 



PARTIAL DIFFERENTIATION; EXPLICIT 99 

df=R(r, s,t,-- -)dr + S(r, s, t, ■ ■ -)ds + T(r, s,t,-- -)dt + ■ • • (20) 

for the total differential in which r, s, t, ■ ■ ■ are Independent variables, 
the coefficients R, S, T, ■ ■ ■ are the derivatives 

R = d -l, S = d -f, T=%,.-: (21) 

cr cs ct v 7 

For in the equation df= Rdr + Sds + Tdt + • • • =f;.dr+f s ds +f t dt -\ , 

the variables r, s, t, • • -, being independent, may be assigned increments 
absolutely at pleasure and if the particular choice dr= 1, ds = dt = --- = 0, 
be made, it follows that R — f r \ and so on. The single equation (20) is 
thus equivalent to the equations (21) in number equal to the number of 
the independent variables. 

As an example, consider the case of the function tan- 1 (y/x). By the rules (19), 

7 ta _ i V _ d (y/g) _ d y/ x — ydx/x 2 _ xdy — ydx . 
x~l+ (y/x) 2 ~ 1 + (y/x) 2 ' x 2 + y 2 

Then — tan- ^ V - = V - — , — tan- 1 V - = — - — , bv (20)-(21). 

ex x x 2 + y 2 cy x x 2 + y- 

If y and x were expressed as y = sinh rst and x — cosh rst, then 

_i V _ xd V — ydx _ [stdr + rtds + rsdt] [cosh 2 rs£ — smh 2 rs£] 



and 



x x 2 + y 2 cosh 2 rs£ + sinh 2 rs£ 

cf st cf rt cf rs 



dr cosh 2 rst cs cosh 2 rst ct cosh 2 rst 



EXERCISES 

1 . Find the partial derivatives f' x . f' or /^, /' f' z of these functions : 
(a) \og(x 2 + y 2 ). (j8) e* cosy sin z, (7) x 2 + 3 xy + y 8 , 



v '■* + ■» 

<„) sin- 1 ^ 




(f) log (sin 
(1) tanh- 1 





^/- / ay + yz + za xs _ 
\x 2 + y 2 + z 2 / 

2. Apply the definition (2) directly to the following to find the partial deriva- 
tives at the indicated points : 

(a) J^- at (1, 1), (/3) x 2 + Sxy + y 3 at (0, 0). and (7) at (1. 1), 

x -t y 

(5) ^-— ^ at (0, 0); also try differentiating and substituting (0, 0). 

3. Find the partial derivatives and hence the total differential of : 



e x, J 



( a ) ~r~, — £' O 3 ) zlogyz, (7) Vfl 2 - x 2 - y 2 . 

x ~ + y / TV \ 

(5) e-*siny, (e) e z2 sinhxy, (f) logtan/jc + -y); 



yr /m x - y 



<<> I • wi+i- (oiogi^ + \/i + 



3x L . z 2 






100 DIFFERENTIAL CALCULUS 

4. Find the general equations of the tangent plane and normal line to these 
surfaces and find the equations of the plane and line for the indicated (x , y ) : 

(a) the helicoid z = fctan- 1 (y/*), (1, 0), (1, - 1), (0, 1), 

(j8) the paraboloid 4pz = (x 2 + y 2 ), (0, p), (2p, 0), (p, — p), 

(7) the hemisphere z = Va 2 — x 2 — ?/ 2 , (0, — | a), (i a, a a), (| V3 a, 0), 

(5) the cubic xyz = 1, (1, 1, 1), (- |, - J, 4), (4, 1, i). 

5. Find the derivative with respect to t in these cases by (10) : 

(a) f = x 2 + if-, x — a cos t, y = 6 sin £, (/3) tan- x -*/- , 2/ = cosh t, * = sinh i, 
(7) sin- 1 (x — ?/), x = St, y = 4t 3 , (S) cos2xy,x = t&n.- 1 t,y = cot- 1 t. 

6. Find the directional derivative in the direction indicated and obtain its 
numerical value at the points indicated : 

(a) x-y, t = 45°, (1, 2), (/3) sin 2 x?/, r = 60 5 , (V3, - 2). 

7. (a) Determine the maximum value of df/ds from (13) by regarding r as 

variable and applying the ordinary rules. Show that the direction that gives the 

maximum is , , 

r = to-4 and then V = J(*J)\(*J)\ 
f x dn \\dx)^\dy) 

(jS) Show that the sum of the squares of the derivatives along any two perpen- 
dicular directions is the same and is the square of the normal derivative. 

8. Show that (/ x ' + y'f' y ) /Vl + V" 1 and (f^if -f^/Vl + y' 2 are the deriva- 
tives of / along the curve y = (x) and normal to the curve. 

9. If df/dn is defined by the work of Ex. 7 (a), prove (14) as a consequence. 

10. Apply the formulas for the change of variable to the following cases : 

(a) , = V^T^,0 = tan-^. Find *, 8 -t, X (W^WX • 

v « x dx cy Wax/ \dy/ 

(13) x = r cos 0, y = r sin 0. Find ^ , ^ , /— Y + — (— Y • 

dr 50 \or/ r 2 \d(p/ 

(7) x = 2 r - 3 s + 7, y = - r + 8 s - 9. Find — = 4x + 2 y if u = x 2 - if. 

dr 

fx = x'cos«-y'sina, ^ow (^ + (^ = (*)*+ (*)*. 

iy = x sm a + y' cos a. \dx/ \cyj \dx'/ \S#7 

( e ) Prove °4- + — = if /(w, ») =f(x-y,y- x). 

ex cy 

(f) Let x = ax' + &y' + cz', ?/ = a'x' + &y + cV, z — a"x' + &'V + c " z ' ■> where 
a, 6, c, a 7 , 6', c', a'', b", c" are the direction cosines of new rectangular axes with 
respect to the old. This transformation is called an orthogonal transformation. Show 

©''©'♦©•= '<&+ ©'+©•= ©"■ 

11. Define directional derivative in space ; also normal derivative and estab- 
lish (14) for this case. Find the normal derivative of / = xyz at (1, 2, 3). 

12. Find the total differential and hence the partial derivatives in Exs. 1, 3, and 
(a) log (x 2 + y 2 + z 2 ), (/3) ?//x, (7) x 2 ?/e^ 2 , (5) xyzlogxyz, 



PAETIAL DIFFERENTIATION; EXPLICIT 101 

(e) u = x' 2 — y 2 , x = rcosst, y = ssinrt. Find cu/cr. cit/cs, cu/dt. 

( $■) u — y/x, x = r cos <p sin 0, y = r sin <f> sin 6. Find u r ', m/, i^ . 
(77) u = e^, x = log Vr 2 + s 2 , y = tan- * (s/r). Find w r ', u/. 

1 o t* cf eg cf eg cf 1 eg 1 3/ 3gr . , . 

13. If — = — and — = , show — = - — and -— = if r, <f> are polar 

ex cy cy ex cr r c<p r c<p or 

coordinates and /, g are any two functions. 

14 . If p (x, y, z, t) is the pressure in a fluid, or p (x, y, z. t) is the density, depend- 
ing on the position in the fluid and on the time, and if u, v, w are the velocities of 
the particles of the fluid along the axes, 

dp cp , cp ep , dp dp cp ep dp. cp 

— = m— + v— + w— + — and -t- = u — + <o -*- + w -4 -\ — - • 
dt ex cy ez ct dt ex cy cz dt 

Explain the meaning of each derivative and prove the formula. 

15. If z = xy, interpret z as the area of a rectangle and mark d x z. A y z. Az on the 
figure. Consider likewise u = xyz as the volume of a rectangular parallelepiped. 

16. Small errors. If /(x, y) be a quantity determined by measurements on x 
and y, the error in / due to small errors dx, dy in x and y may be estimated as 
df = f' x dx + f' y dy and the relative error may be taken as df -i-f = d log/. Why 
is this ? 

(a) Suppose £ = | ab sin C be the area of a triangle with a = 10, b = 20, C = 30°. 
Find the error and the relative error if a is subject to an error of 0.1. Ans. 0.5, 1%. 
(j3) In (a) suppose C were liable to an error of 10' of arc. Ans. 0.27, \%. 

(7) If a, 6, C are liable to errors of 1%, the combined error in S may be 3.1%. 

(5) The radius r of a capillary tube is determined from 13.6 7rf 2 l = w by find- 
ing the weight w of a column of mercury of length I. If w = 1 gram with an error 
of 10- 3 gr. and I = 10 cm. with an error of 0.2 cm., determine the possible error 
and relative error in r. Ans. 1.2%, 6 x 10- 4 , mostly due to error in I. 

( e ) The formula c 2 = a 2 + b 2 — 2 ab cos C is used to determine c where a = 20, 
b = 20, C = 60° with possible errors of 0.1 in a and b and 30' in C. Find the possible 
absolute and relative errors inc. Ans. 1, 1\%. 

(f) The possible percentage error of a product is the sum of the percentage 
errors of the factors. 

( rj ) The constant g of gravity is determined from g = 2 st~ 2 by observing a body 
fall. If s is set at 4 ft. and t determined at about | sec, show that the error in g 
is almost wholly due to the error in t, that is, that s can be set very much more 
accurately than t can be determined. For example, find the error in t which would 
make the same error in g as an error of | inch in s. 

(6) The constant g is determined by gt 2 = tt 2 1 with a pendulum of length I and 
period t. Suppose t is determined by taking the time 100 sec. of 100 beats of the 
pendulum with a stop watch that measures to 4 sec. and that I may be measured 
as 100 cm. accurate to \ millimeter. Discuss the errors in g. 

17. Let the coordinate x of a particle be x =f(q v q 2 ) and depend on two inde- 
pendent variables g 1? q 2 . Show that the velocity and kinetic energy are 



102 DIFFERENTIAL CALCULUS 

where dots denote differentiation by t, and a u , a 12 , a 22 are f mictions of (g t , q 2 ). 

Show — = — , i = 1, 2, and similarly for any number of variables q. 
' m dqi 

18. The helix x = a cost, y — a suit, 2 = Han a: cuts the sphere x 2 + y 2 + z 2 = 
a 2 sec 2 /3 at sin~ * (sin a sin /3) . 

19. Apply the Theorem of the Mean to prove that /(x, y, z) is a constant if 
f' x = f y =f z — is true for all values of x, ?/, z. Compare Theorem 16 (§ 27) and 
make the statement accurate. 



20. Transform — = -J(§Q + (— ) + (?) t0 ( a ) cylindrical and (/3) polar 
coordinates (§ 40). 

21. Find the angle of intersection of the helix x = 2cosi, y = 2sin£, z = t and 
the surface xyz = 1 at their first intersection, that is, with < t < \ -rr. 

22. Let/, g, h be three functions of (x, y, z). In cylindrical coordinates (§ 40) 
form the combinations F = /cos 4> + g sin <p, G = — /sin <p -{- g cos 0, Zf = A. Trans- 

?X C?/ C2 C>^ 52 dX cty 

to cylindrical coordinates and express in terms of F, G, R in simplest form. 

23. Given the functions y x and {zv) x and z^ x \ Find the total differentials and 
hence obtain the derivatives of x x and (x x ) x and x( xX \ 

50. Derivatives of higher order. If the first derivatives be again 
differentiated, there arise four derivatives f^, f xy , f yx , f' y ' y of the second 
order, where the first subscript denotes the first differentiation. These 
may also be written 

where the derivative oi cf/dy with respect to as is written d 2 f/8xdy 
with the variables in the same order as required in D x D y f and opposite 
to the order of the subscripts in f yx . This matter of order is usually of 
no importance owing to the theorem : If the derivatives f' x , f have 
derivatives f xy , f yx which are continuous in (x, y) in the neighborhood 
of any -point (x Q , y ), the derivatives f xy and f yx are equal, that is, 

The theorem may be proved by repeated application of the Theorem of the 
Mean. For 

[/(x + h, y + k)-f(x , y + k)]- [/(x + h, y )-f(x , y )] = [</>{y + k)-<t>(y )J 
= [/(x + h, y + k)-f(x + h, y )] - [/(x , y + k)-f(x , y )] = ty(x + h)-^(x )} 

where <p(y) stands for /(x + h, ?y)-/(x , y) and ^(x) for /(x, y + k)-f(x, y ). 

Now 

<f>(y + k) - <f>{y ) = k<p'(y + 0k) = k[f y {x + h, y + 0k) -f' y {x , y + 6k)], 
1// (x + h)-f (x ) = h+'(x + B'h) = h[f x (x + 6% y + k) - f x (x + 6% y )J 



PARTIAL DIFFERENTIATION; EXPLICIT 10a 

by applying the Theorem of the Mean to <p (y) and \f/ (x) regarded as functions of a 
single variable and then substituting. The results obtained are necessarily equal 
to each other ; but each of these is in form for another application of the theorem. 

k [fy( X + h i Vo + 0k ) ~fy( X V Vo + &k )1 = kh fyx( X + V h , V + & k ), 
Hfx( X + 0% Vo + k ) ~fx( X + 0% Vo)l = hk f"y( X + #% VO + -n' k )- 

Hence f yx (x + nth, y + 6k) = f xy (x + 0% y + ri'k). • 

As the derivatives f yx , f xy are supposed to exist and be continuous in the variables 
(x, y) at and in the neighborhood of (x , y ), the limit of each side of the equation 
exists as h = 0, k = and the equation is true in the limit. Hence 

fyx \ X Qi Vo) = fxy \ X Qi Vq) ' 

The differentiation of the three derivatives f^, f xy = f yx , f' yy will give 
six derivatives of the third order. Consider f xxy and f xyx . These may 
be written as (f^)^ and (f' x )' y ' x and are eqnal by the theorem just proved 
(provided the restrictions as to continuity and existence are satisfied).. 
A similar conclusion holds for f yxy and f yyx ; the number of distinct 
derivatives of the third order reduces from six to four, just as the 
number of the second order reduces from four to three. In like manner 
for derivatives of any order, the value of the derivative depends not on 
the order in which the individual differentiations ivith respect to x and 
y are performed, but only on the total number of differentiations with 
respect to each, and the result may be written with, the differentiations 
collected as 



2>«I>« f = -7— A = fZ I n \ etc. (22) 



c m + n f 
dx m dy T 

Analogous results hold for functions of any number of variables. If 
several derivatives are to be found and added together, a symbolic 
form of writing is frequently advantageous. For example, 

ex oycz cy 

(P. + D s f/ = {pi + 2 d x d v + D$f =/z + 2/;;+ /;;. 

51. It is sometimes necessary to change the variable in higher deriv- 
atives, particularly in those of the second order. This is done by a 
repeated application of (18). Thus f" r would be found by differentiat- 
ing the first equation with respect to r, and f£ by differentiating the 
first by s or the second by r, and so on. Compare p. 12. The exercise 
below illustrates the method. It may be remarked that the use of higher 
differentials is often of advantage, although these differentials, like the 
higher differentials of functions of a single variable (Exs. 10, 16-19, 
p. 67), have the disadvantage that their form depends on what the 
independent variables are. This is also illustrated below. It should be 
particularly borne in mind that the great value of the first differential 



104 DIFFEBENTIAL CALCULUS 

lies in the facts that it may be treated like a finite quantity and that 
its form is independent of the variables. 

To change the variable in v^. + v' y ' y to polar coordinates and show 
d 2 v d 2 v d 2 v 1 dv 1 d 2 v f x — r cos 0, y = r sin 0, 



ex 



dy 2 • dr 2 r dr r 2 d<p 2 {_ r = Vx 2 + y 2 , <p = tsm- 1 (y/x). 



dv dv dr dv d<f> dv dv dr dv d<b 

Then — = + , — = \- - 

dx dr dx d<p dx dy or dy d<p dy 

by applying (18) directly with x, y taking the place of r, s, • • • and r, <p the place 
of x, ?/, z, • • • . These expressions may be reduced so that 

dv _ dv x dv — y _dvx dv — y 

dx dr Vx 2 + y 2 ^0 x ~ + V 2 ° r r d<P r2 
d 2 v _ d dv _ d dv dr d dv dep 
dx 2 dx dx dr dx dx d<p dx dx 

[d 2 v x dv d x d 2 v — y dv d — ylx 
dr 2 r dr dr r drccp r 2 d<f> dr r 2 J r 

[d 2 v x dv d x d 2 v — y dv d — y~\— y 
d4>dr r dr dcf> r d<f> 2 r 2 dcf> dcf> r 2 J r 2 

The differentiations of x/r and — y/r 2 may be performed as indicated with respect to 
r, 0, remembering that, as r, are independent, the derivative of r by is 0. Then 

d 2 v _ x' 2 d 2 v y 2 dv xy d 2 v xy dv y 2 d 2 v 

dx 2 r 2 dr 2 r 3 dr r 3 drd<f> i A d<p r 4 d<p 2 

In like manner d 2 v/dy 2 may be found, and the sum of the two derivatives reduces 
to the desired expression. This method is long and tedious though straightforward. 
It is considerably shorter to start with the expression in polar coordinates and 
transform by the same method to the one in rectangular coordinates. Thus 



dv dv dx dv dy dv dv . 

— = 1 = — : COS 0-1 Sill 

dr dx dr dy dr dx dy 



1/dv dv \ 

-\t x + tv)' 

r\dx dy J 

d I dv\ /d 2 v "" , d 2 v .-- . \ , / d 2 v d 2 v . \ -dv ^ , dv . \ 

— (r — ) = ( — cos0 H sm0 )x + -cos0 -\ sin0 )y -\ cos0 -\ -sm0, 

dr\ dr] \dx 2 dydx V \dxdy dy 2 J dx * dy 



dv dv dx dv dy dv . dv dv dv 

— = 1 • — = r sm H r cos = y ^ x, 

dep dx d<p dy d(f> dx dy dx dy 



ld 2 v . d 2 v \ I d 2 v . d 2 v \ 

— sm cos ) y + ( sm <t> -\ cos J x 

\dx 2 dydx J \ dxdy dy 2 J 



dv dv . 

cos sm 0. 

dx dy 

1 d 2 v ld 2 v d 2 v\ 
Then — r— + , = — - + -— \r 



A/V^Ui— = (— + — ) 

dr\ dr) r d<j> 2 \dx 2 dy 2 ] 



d 2 v d 2 v Id/ dv\ 1 d 2 v d 2 v ldv 1 d 2 v 

or = ir — I H = 1 (23) 

ax 2 dy 2 rdr\ dr) r 2 d(p 2 dr 2 r dr r 2 dcf> 2 

The definitions d 2 f = f^dx 2 , d x d y f = f^dxdy, d 2 f — f' y ' y dy 2 would naturally be 
given for partial differentials of the second order, each of which would vanish if / 
reduced to either of the independent variables x, y or to any linear function of 
them. Thus the second differentials of the independent variables are zero. The 



PARTIAL DIFFERENTIATION; EXPLICIT 105 

second total differential would be obtained by differentiating the first total differ- 
ential. 

d 2 f = ddf = d( C -^dx + — dy) = d—dx + d — dy + ^-d 2 x + ^-d 2 y ; 
\dx dy ) ex dy ex dy 

df d 2 f d 2 f df d 2 f d 2 f 

but d— = — dx + -^-dy, d^- = -^-dx + —dy, 

dx ex 2 cycx cy dxdy cy 2 

and d 2 f = C -Ldx 2 + 2 — dxdy + — dy 2 + ^ d 2 x + — d 2 y. (24) 

ex 2 dxdy dy 2 dx cy 

The last two terms vanish and the total differential reduces to the first three terms 
if x and y are the independent variables ; and in this case the second derivatives, 
fxxi fxyi fyyi are tne coefficients of dx 2 , 2 dxdy, dy 2 , which enables those derivatives 
to be found by an extension of the method of finding the first derivatives (§ 49). 
The method is particularly useful when all the second derivatives are needed. 
The problem of the change of variable may now be treated. Let 

d 2 t) d 2 v r 2 V 

d 2 v = — dx 2 + 2 — dxdy + — dy 2 
ex 2 dx 2 dy 2 

d 2v > ,i •? . ^ d 2 v , , , d 2 v dv , 9 , dv ., 

= — dr 2 + 2 drd(b + — dd> 2 + — d 2 r + — d 2 <p, 

dr 2 drd<p d<p 2 dr c</> 

where x, y are the independent variables and r, <p other variables dependent on 
them — in this case, defined by the relations for polar coordinates. Then 

dx = cos <pdr — r sin 0d0, dy = sin <f>dr + r cos <pd<f> 
or dr = cos cf>dx + sin <f>dy, rdcp = — sin cpdx + cos <pdy. (25) 

Then d 2 r = (— sin <pdx + cos <pdy) dcj> = rd<j>d<p = rdcp 2 , 

drdcp + rd 2 <p = — (cos <pdx + sin <pdy) d(f> = — drdcp, 

where the differentials of dr and rdcp have been found subject to d 2 x = d 2 y = 0. 
Hence d 2 r = rdcp 2 and rd 2 cp = — 2drd<p. These may be substituted in d 2 v which 
becomes 

>/ \d(p 2 dr) 

Next the values of dr 2 , drdcp, d<p 2 may be substituted from (25) and 



dr 2 \crc0 r c0/ 



,„ r 8 « 9 2 /^ 1M • /c' 2 v a?Asin 2 0l 79 

d 2 v = — - cos 2 cos d> sm + ( h r — \ dx 2 

Idr 2 * r\drd<p rdcp) W' 2 dr) r 2 J 

, _ Tc 2 u . / c 2 u 1 cu\ cos 2 — sin 2 c 2 u cos sin 0~| , _ 

+ 2 — cos sm + T. Z . *L — Z dxdy 

\_cr 2 \ordcp r dep) r d<p 2 r 2 J 

, VcH . _ , 2/d 2 v 1 dv\ . /d 2 v cu\cos 2 0l^ . 

+ ^sm 2 + -(-— - - — cos0sm0 + — - + r-A—f-W. 
\_dr 2 r \drccp r dep) \dcp 2 dr) r 2 J 

Thus finally the derivatives v xx , v xy , v£ are the three brackets which are the 
coefficients of dx 2 , 2 dxdy, dy 2 . The value of v xx + v^ is as found before. 

52. The condition f' x ' y = f£ which subsists in accordance with the 
fundamental theorem of § 50 gives the condition that 

cf cf 

M(x, y) dx + N(x, y) dy = ■£- dx + -£- dy = df 






106 DIFFERENTIAL CALCULUS 

be the total differential of some function f(x, y). In fact 

c cf _ dM _ dX _ c cf 
cy dx cy dx ex dy 
dM dX /dM\ A/A 



dy dx ° } \dy) x \dx) y 

The second form, where the variables which are constant during the 
differentiation are explicitly indicated as subscripts, is more common in 
works on thermodynamics. It will be proved later that conversely if 
this relation (26) holds, the expression Mdx + Xdy is the total differ- 
ential of some function, and the method of finding the function will 
also be given (§§ 92, 124). In case Mdx + Xdy is the differential of 
some function f(x, y) it is usually called an exact differential. 

The application of the condition for an exact differential may be 
made in connection with a problem in thermodynamics. Let S and U 
be the entropy and energy of a gas or vapor inclosed in a receptacle of 
volume v and subjected to .the pressure p at the temperature T. The 
fundamental equation of thermodynamics, connecting the differentials 
of energy, entropy, and volume, is 

dU=TdS-pdv, and (^-) = - (±) (27) 



\dv) 8 \dS 

is the condition that dU be a total differential. Now, any two of the 
five quantities U, S, v, T, p may be taken as independent variables. In 
(27) the choice is S, v; if the equation were solved iov clS, the choice 
would be U, v ; and U, S if solved for dv. In each case the cross differ- 
entiation to express the condition (26) would give rise to a relation 
between the derivatives. 

If p, T were desired as independent variables, the change of variable 

should be made. The expression of the condition is then 

\df[ \dp)r P \dp) t] J P ~ \dpl [dTjp V \dTj P \)T 
/dS\ cS c-v _ c 2 S t dv \ c 2 v 

\dp) T cTcp P cTcp~ cpcT \dT/ p P cpcT' 

where the differentiation on the left is made with p constant and that on the right 
with T constant and where the subscripts have been dropped from the second 
derivatives and the usual notation adopted. Everything cancels except two terms 
which oive 



PARTIAL DIFFERENTIATION ; EXPLICIT 107 

m=-(*L) or It™*) =-(!*)■.' (28) 

\dp)r \dTjp T\dp/T \dTj P y ' 

The importance of the test for an exact differential lies not only in the relations 
obtained between the derivatives as above, but also in the fact that in applied 
mathematics a great many expressions are written as differentials which are not 
the total differentials of any functions and which must be distinguished from exact 
differentials. For instance if dH denote the infinitesimal portion of heat added 
to the gas or vapor above considered, the fundamental equation is expressed as 
dH = dll + pdv. That is to say, the amount of heat added is equal to the increase 
in the energy plus the work done by the gas in expanding. Now dH is not the dif- 
ferential of any function H(U, v) ; it is dS = dH/T which is the differential, and 
this is one reason for introducing the entropy S. Again if the forces X, Y act on a 
particle, the work done during the displacement through the arc ds — V ' dx' 1 + dy' 2 
is written dW = Xdx + Ydy. It may happen that this is the total differential of 
some function ; indeed, if 

d W = - dV(x, y), Xdx + Ydy = - dV, X = - — , Y=-—, 

ex cy 

where the negative sign is introduced in accordance with custom, the function V is 
called the potential energy of the particle. In general, however, there is no poten- 
tial energy function V, and d W is not an exact differential ; this is always true 
when part of the work is due to forces of friction. A notation which should dis- 
tinguish between exact differentials and those which are not exact is much more 
needed than a notation to distinguish between partial and ordinary derivatives ; 
but there appears to be none. 

Many of the physical magnitudes of thermodynamics are expressed as deriva- 
tives and such relations as (26) establish relations between the magnitudes. Some 
definitions : 

specific heat at constant volume is C v = ( — I = T I — ) , 

\dTJv \dTJv 

specific heat at constant pressure is C D = ( ) = Tl — ) 

latent heat of expansion 

coefficient of cubic expansion is a p = - I — ) 

modulus of elasticity (isothermal) is E T = — 

modulus of elasticity (adiabatic) is Eg = — vl — ) 

\dv/s 



\dv Jt 



„/dS\ 
dv/'j 



53. A polynomial is said to be homogeneous when each of its terms 
is of the same order when all the variables are considered. A defini- 
tion of homogeneity which includes this case and is applicable to more 
general cases is : A function f(x, y, z, • • •) of any number of variables is 
called homogeneous if the function is multiplied by some power of X tvhen 
all the variables are multiplied by A.; and the power of X which factors 



108 DIFFERENTIAL CALCULUS 

out is called the order of homogeneity of the function. In symbols the 
condition for homogeneity of order n is 

f(Xx, Xy, Xz, ■•■) = X«f(x, y,z,.. •). (29) 

Thus x£ + £, ^f + tan- 1 -, - 1 (29') 

x z l z ^/x 2 + y 2 

are homogeneous functions of order 1, 0, — 1 respectively. To test a 
function for homogeneity it is merely necessary tb replace all the vari- 
ables by X times the variables and see if X factors out completely. The 
homogeneity may usually be seen without the test. 

If the identity (29) be differentiated with respect to X, 

\ x dx" iry dJj JrZ dz + " jf( Xx > Xy > Xz > ' ' ■) = nXn ~ x f^ y> z >-- ■)■ 

A second differentiation with respect to X would give 



.2 ^ _ 



/ c 2 d 2 d 2 \ 

° r I* 2 dx 1 + 2xy dxty + y2 df + * " ') f= n(jl ~ ^ Xn ~ 2f ( x > y \*>" ')■ 

Now if X be set equal to 1 in these equations, then 

In words, these equations state that the sum of the partial derivatives 
each multiplied by the variable with respect to which the differentia- 
tion is performed is n times the function if the function is homogeneous 
of order n ; and that the sum of the second derivatives each multiplied 
by the variables involved and by 1 or 2, according as the variable is 
repeated or not, is n (n — 1) times the function. The general formula 
obtained by differentiating any number of times with respect to X may 
be expressed symbolically in the convenient form 

(xD x + yD y + zD z + .- .)*/= n(n-T)---(n-k + 1)/. (31) 

This is known as Eider's Formula on homogeneous functions. 

It is worth while noting that in a certain sense every equation which represents 
a geometric or physical relation is homogeneous. For instance, in geometry the 
magnitudes that arise may be lengths, areas, volumes, or angles. These magni- 
tudes are expressed as a number times a unit ; thus, V2 ft., 3 sq. yd., ir cu. ft. 



PAETIAL DIFFERENTIATION; EXPLICIT 109 

In adding and subtracting, the terms must be like quantities ; lengths added to 
lengths, areas to areas, etc. The fundamental unit is taken as length. The units of 
area, volume, and angle are derived therefrom. Thus the area of a rectangle or 
the volume of a rectangular parallelepiped is 

A = a ft. x b ft. - ah ft.' 2 = ab sq ft,, V = a ft. x b ft. x c f t. = abc ft, 3 = ahc cu. ft., 

and the units sq. ft., cu. ft. are denoted as ft. 2 , ft. 3 just as if the simple unit ft. 
had been treated as a literal quantity and included in the multiplication. An area 
or volume is therefore considered as a compound quantity consisting of a number 
which gives its magnitude and a unit which gives its quality or dimensions. If L 
denote length and [L] denote "of the dimensions of length, "' and if similar nota- 
tions be introduced for area and volume, the equations [A] = [L] 2 and [I"] = [X] 3 
state that the dimensions of area are squares of length, and of volumes, cubes of 
lengths. If it be recalled that for purposes of analysis an angle is measured by the 
ratio of the arc subtended to the radius of the circle, the dimensions of angle are 
seen to be nil, as the definition involves the ratio of like magnitudes and must 
therefore be a pure number. 

When geometric facts are represented analytically, either of two alternatives is 
open : 1°, the equations may be regarded as existing between mere numbers ; or 
2°, as between actual magnitudes. Sometimes one method is preferable, sometimes 
the other. Thus the equation x 2 + y 2 = r 2 of a circle may be interpreted as 1°, the 
sum of the squares of the coordinates (numbers) is constant ; or 2°, the sum of the 
squares on the legs of a right triangle is equal to the square on the hypotenuse 
(Pythagorean Theorem). The second interpretation better sets forth the true 
inwardness of the equation. Consider in like manner the parabola y 2 = 4_px. Gen- 
erally y and x are regarded as mere numbers, but they may equally be looked 
upon as lengths and then the statement is that the square upon the ordinate equals 
the rectangle upon the abscissa and the constant length 4p ; this may be inter- 
preted into an actual construction for the parabola, because a square equivalent 
to a rectangle may be constructed. 

In the last interpretation the constant p was assigned the dimensions of length 
so as to render the equation homogeneous in dimensions, with each term of the 
dimensions of area or [L] 2 . It will be recalled, however, that in the definition of 
the parabola, the quantity p actually has the dimensions of length, being half the 
distance from the fixed point to the fixed line (focus and directrix). This is merely 
another corroboration of the initial statement that the equations which actually 
arise in considering geometric problems are homogeneous in their dimensions, and 
must be so for the reason that in stating the first equation like magnitudes must 
be compared with like magnitudes. 

The question of dimensions may be carried along through such processes as 
differentiation and integration. For let y have the dimensions [y] and x the dimen- 
sions [x]. Then Ay, the difference of two y's, must still have the dimensions [y~\ 
and Ax the dimensions [x]. The quotient Ay /Ax then has the dimensions |j/]/[x]. 
For example the relations for area and for volume of revolution, 



dA dV _ . [dAl [A] rT1 rdYl [F] rT12 

= nf, give L_J = LJ = [i]f |_J = L_i = [i]2i 

and the dimensions of the left-hand side check with those of the right-hand side. 



dx 7 dx ' ■ LdxJ [X] u J7 Idxl [X] 

tie dimensions of the left-hand side check with those of the rig 
As integration is the limit of a sum, the dimensions of an integral are the product 



110 



DIFFERENTIAL CALCULUS 



of the dimensions of the function to be integrated and of the differential dx. 
Thus if 

y — 



r x dx 1 ,i 

= / 2 ■ 2 = _ ^ - + 

Jo a 2 + x l a a 



were an integral arising in actual practice, the very fact that a 2 and x 2 are added 
would show that they must have the same dimensions. If the dimensions of x 
be [£], then 



and this checks with the dimensions on the right which are [I*] -1 , since angle has 
no dimensions. As a rule, the theory of dimensions is neglected in pure mathe- 
matics ; but it can nevertheless be made exceedingly useful and instructive. 

In mechanics the fundamental units are length, mass, and time ; and are denoted 
by [-L], [M~\, [T], The following table contains some derived units : 



velocity 
areal velocity 
angular velocity 



m 

[T]' 
[T] 

i 
m' 



acceleration 



density 



moment 



m 

m 2 ' 
m 

mm* 
[T]2 ■ 



force 



momentum 



energy 



[M] [L] 

[T]2 



With the aid of a table like this it is easy to convert magnitudes in one set of 
units as ft., lb., sec, to another system, say cm., gm., sec. All that is necessary is 
to substitute for each individual unit its value in the new system. Thus 

ft. 



321 



1 ft. = 30.48 cm., g = 321 x 30.48 



cm. 
sec. 2 



EXERCISES 

1. Obtain the derivatives/^, /^, /^, f' y ' y and verify/^ = f yx . 



(a) sin 



OS) lo 



xyi J yx 

x 2 + v 2 



xy 



( 7 ) <p(^\ + t(xy), 



2. Compute d 2 v/dy 2 in polar coordinates by the straightforward method. 

c 2 v B 2 v 

3. Show that a 2 — = — if v =f(x + at) + (x — at). 

4. Show that this equation is unchanged in form by the transformation 



d 2 f 



+ 2xy 2 ^- + 2(y-y*)^ + x 2 y 2 f=0; u = xy, v = l/y. 
dx 2 dx dy 

5. In polar coordinates z = r cos #, x = r sin 6 cos 0, y = r sin 6 sin in space 

dx 2 dy 2 dy 2 r 2 ldr\ drj sin 2 0dcp 2 smOd0\ c6j\ 
The work of transformation may be shortened by substituting successively . 
x = r 1 cos 0, y = r x sin 0, and z = r cos 0, r i = r sin 0. 

6. Let x, 2/, 2, t be four independent variables and x = r cos <p, y = r sin <p, z = z 
the equations for transforming x, 2/, z to cylindrical coordinates. Let 



PAKTIAL DIFFERENTIATION; EXPLICIT 111 





X = 


8 2 / 

cxdz 


F = 


cycz' 


Z = 


ex 2 a?/ 2 


, F = 


cyc£ 


, G = 


dxdt 


show 


z = 


.iaQ } 

r cr 


Xcos 


+ Fsin 


= 


_iaQ } 

r dz 


F si n0 


- G 


COS0 = 


r dt ' 



where r- r Q = df/dr. (Of importance for the Hertz oscillator.) 

7. Apply the test for an exact differential to each of the following, and write 
by inspection the functions corresponding to the exact differentials : 

(a) Sxdx + y' 2 dy, (/3) Sxydx + x s dy, (y) x 2 ydx + y 2 dy, 

xdx + ydy xdx — ydy ydx — xdy 

{8) x 2 + y 2 ' (6) x 2 + y 2 ' (f) x 2 + y 2 ' 

(t?) (4 x 3 + 3 x 2 y + y 2 ) dx + (x 3 + 2 xy + 3 y 3 ) dy, (0) x 2 y 2 (dx + dy). 

8. Express the conditions that P(x, y, z)dx + Q(x, y, 2)dy + R(x, y, z)dz be 
an exact differential dF(x, y, z). Apply these conditions to the differentials : 

(a) 3 x 2 y 2 zdx + 2 x s yzdy + x 3 y 2 dz, (p) (y + z) dx + (x + z) dy + (x + y) dz. 

9. Obtain (— — ) = ( — ] and ( — ) = ( — ] from (27) with proper variables. 

\dT) v \dv) T \dS) p \dp) s 7 

10. If three functions (called thermodynamic potentials) be defined as 

xp = U-TS, x=U + pv, £=U-TS+pv, 

show dxfy = - SdT - pdv, d x = TdS + vdp, d{= - SdT + vdp, 

and express the conditions that d\p, dx, d£ be exact. Compare with Ex. 9. 

11. State in words the definitions corresponding to the defining formulas, p. 107. 

12. If the sum (Mdx + Ndy) + (Pdx + Qdy) of two differentials is exact and one 
of the differentials is exact, the other is. Prove this. 

13. Apply Euler's Formula (31), for the simple case k = 1, to the three func- 
tions (29') and verify the formula. Apply it for k = 2 to the first function. 

14. Verify the homogeneity of these functions and determine their order : 

xyz 



(a) y 2 /x + x(logx — logy), 


IP) " ■ 


(7) 




Vx 2 + y 2 




(5) xye^ + z 2 , 


(e) Vxcot- 1 -, 
v ' z 


(f) 



ax + by + cz 
Vx — Vy _ 
</x + </y 

15. State the dimensions of moment of inertia and convert a unit of moment of 
inertia in ft. -lb. into its equivalent in cm.-gm. 

16. Discuss for dimensions Peirce's formulas Nos. 93, 124-125, 220, 300. 

1 n r, j.- -n in -i^i x t d dx dv 1 d dT . dx cT 

17. Continue Ex. 1/, p. 101, to show = — and = mv 1 

dt dqi dqi dt din dqi dq x 

18. If pi = — in Ex. 17, p. 101, show without analysis that 2 T — q.p, -f q<>p 9 . 

dqi 

If T denote T" = T, where T' is considered as a function of p^ p while T is con- 
sidered as a function of q v g 2 , prove from T' = q l p l + q 2 P 2 — T that 

dT _ . dT _ _dT 

dpi l dqi dqi 



112 DIFFERENTIAL CALCULUS 

19. If (x 15 2/j) and (x 2i y 2 ) are the coordinates of two moving particles and 

m ^l = X t , m l ^ = Y 1 , m 2 d ^ = X„, m 2 ^ = y 
1 dt 2 v l dt 2 *' 2 dt* 2 ' 2 dZ 2 2 

are the equations of motion, and if x v y v x 2 , y 2 are expressible as 

X l =/l(ffll 32? 0s)> Kl = 01 fall 02> %) > X 2 =/ 2 fe 02^ 3 )> 02 = QiiQv %, %) 

in terms of three independent variables q x , g 2 , g 3 , show that 

Q Y ^ +F ^ +X ^2 + r ^2 = l^_^, 

^ 1 a ?1 x e 3l 2 a gi 2 ag x efta^ ag x 

where T= £ (m^ 2 + m 2 w 2 2 ) = T(g r , g 2 , # 3 , g l5 g 2 , # 8 ) and is homogeneous of the 
second degree in q v q 2 , q s . The work may be carried on as a generalization of 
Ex. 17, p. 101, and Ex. 17 above. It may be further extended to any number of 
particles whose positions in space depend on a number of variables q. 

20. In Ex. 19 if pi = — , generalize Ex. 18 to obtain 

c'qi 

. . _ dT_ dT__dT^ _dp 1 dT_ 

qi ~ dpi' ~dq~i~ ~dq t ' l ~~dt + ~dq~ 1 ' 

The equations Qi = and Qi = — -\ are respectively the Lagran- 

dt cqi dqi dt dqi 

gian and Hamiltonian equations of motion. 

21. If rr' — k 2 and <p' = and vf (r', 0') = v (r, 0), show 

av i w i av _ r 2 /d 2 v ldv 1 d 2 v\ 

cr' 2 r' dr' r' 2 c<p" 2 r' 2 \dr 2 r dr r 2 d<p 2 / 

k 

22. If rr' = k 2 , <j>' = 0, 6' = 0, and v'{r\ <p', 6') — -v(r, 0, #), show that the 

r' 

expression of Ex. 5 in the primed letters is kr 2 /r' z of its value for the unprimed 
letters. (Useful in § 198.) 

23. If z = xJ?) + *(-), show x 2 — + 2xy — + y 2 — = 0. 

W \xj dx 2 dxdy dy 2 

24 . Make the indicated changes of variable : 

. , dW d 2 V 9 (c 2 V d 2 V\ ., 

(a) 1 = e- 2m 1 if x = e u cos v, y = e u sin v, 

dx 2 dy 2 \ du 2 dv 2 1 

w_ aw = ,*r d*v\ r/a/y mh where 

KP) du 2 ^ dv 2 \dx 2 dy 2 )\\du] ^\dvj J 

x=f{u,v), y = $(u,v) y — = — , — = -— • 

du cv dv du 

25. For an orthogonal transformation (Ex. 10 (f), p. 100) 

c 2 v d 2 v d 2 v _ d 2 v d 2 v d 2 v 
dx 2 dy 2 dz 2 dx" 2 dy" 2 dz" 2 

54. Taylor's Formula and applications. The development of f(x, y) 
is found, as was the Theorem of the Mean, from the relation (p. 95) 



PAETIAL DIFFERENTIATION; EXPLICIT 118 

A/=<&(l)-*(0) if *(t)=f(a + th,b + tk). 

If ®(t) be expanded by Maclaurin's Formula to n terms, 

*(0 - *(o> = »i(0) + 1 *» ( o) + . . . + ^^ *<»-d(0) + J *»(«>. 

The expressions for $'(f) and $'(0) may be found as follows by (10) : 

* r (o = w* + w;, *'(o) = [v: + */a, _ ., 

then $"(0 = a (h& + 7./;;) + a- (///;; + /./;) 

= #/£ + 2 /,/./,; + &% = (ad. + i-ojy, 

*»(*) = (AD. + ftD„)y, •»(<>) = [(AD. + **>,)'/]._.. 

And f(a + A, J + k) -f(a, h) = \f = $(1) - *(0) = (hD x + kD,)f(a, b) 
+ A (AD. + fcD„)V(«, 5) + ■ ■ ■ + — ±j-, (AD. + kD v )«-if(a, b) 

+ 1 (AC. + *D r y/(a + <?A, J + Ofc). (32) 

In this expansion, the increments h and A' may be replaced, if de- 
sired, by x — a and y — b and then f(x, y) will be expressed in terms 
of its value and the values of its derivatives at (a, b) in a manner 
entirely analogous to the case of a single variable. In particular if the 
point (a, b) about which the development takes place be (0, 0) the 
development becomes Maclaurin's Formula f or f(x, y). 

/Oh y) =/(o, 0) + (*d. + yD„)/(o, o) + i (xd. + yD v Yf(o, o) + ■ • • 

+ ( re 3l)T" (a: ^ + ^ ) "" 1/(0 ' 0) + J (^H^*)"/^ %)■ (32') 

Whether in Maclaurin's or Taylor's Formula, the successive terms are 
homogeneous polynomials of the 1st, 2d, • • •, (n ■ — l)st order in x, y or 
in x — a, y — b. The formulas are unique as in § 32. 



Suppose V 1 — x 2 — y 2 is to be developed about (0, 0). The successive deriva- 
tives are 

.£= , ~ x => /;= , ~ y =. /;(o,o) = o, /;(o,o) = o, 

VI — x 2 — y 2 V 1 — x 2 — y 2 

„ _ -l + ij 2 „ xy „ - 1 + x 2 

J XX _ 3 ' ■'33/ A ' «'?/»/ 



f ,„ |(l-y 2 )x ,„ y3_2xy 2 -y 

(1 _ £-2 _ ^2)f (1 _ 3.2 _ ^2)1 



and Vl - x 2 - y 2 = 1 '+ (0a> + Oy) + i(- x 2 + Ox?/ - ?/ 2 ) + i(0x 3 + ...) + ..., 
or V 1 — x 2 — ?/ 2 = 1 — ^ (x 2 + ?/ 2 ) + terms of fourth order + • • • . 

In this case the expansion may be found by treating x 2 + y 2 as a single term and 
expanding by the binomial theorem. The result would be 



114 DIFFERENTIAL CALCULUS 

[1 _ (X 2 + ytyfi = 1 _ 1 {X 2 + y2) _ 1 (X 4 + 2 X V + ^ _ j_ (a , 2 + y2)3 _ 

That the development thus obtained is identical with the Maclaurin development 
that might be had by the method above, follows from the uniqueness of the devel- 
opment. Some such short cut is usually available. 

55. The condition that a function z=f(x, y) have a minimum or 
maximum at (a, b) is that Af > or Af < for all values of h = Ax 
and k = Ay which are sufficiently small. From either geometrical or 
analytic considerations it is seen that if the surface z —f(x, y) has a 
minimum or maximum at (a, b), the curves in which the planes y = b 
and x = a cut the surface have minima or maxima at x = a and y = b 
respectively. Hence the partial derivatives f x and f' y must both vanish 
at (a, b), provided, of course, that exceptions like those mentioned on 
page 7 be made. The two simultaneous equations 

/* = 0, /; = 0, (33) 

corresponding to f'(x) = in the case of a function of a single varia- 
ble, may then be solved to find the positions (x, y) of the minima 
and maxima. Frequently the geometric or physical interpretation of 
z = f(x, y) or some special device will then determine whether there 
is a maximum or a minimum or neither at each of these points. 

For example let it be required to find the maximum rectangular parallelepiped 
which has three faces in the coordinate planes and one vertex in the plane 
x/a + y/b + z/c = 1. The volume is 



xyz = cxy 



R-I) 



dV c c cV c c 

— = — 2-xy y 2 + cy = — = — 2 - xy x 2 + ex = 0. 

dx a b cy b a 

The solution of these equations is x = } a, y = ^ b. The corresponding z is ic and 
the volume V is therefore abc/9 or f of the volume cut off from the first octant by 
the plane. It is evident that this solution is a maximum. There are other solutions 
of Y' x = Vy = which have been discarded because they give V = 0. 

The conditions f' x = f' y = may be established analytically. For 

a/ = (/: + Q ^x + (/; + Q A./. 

Now as £ 1? £ 2 are infinitesimals, the signs of the parentheses are deter- 
mined by the signs of f x , f* unless these derivatives vanish ; and hence 
unless f' x = 0, the sign of Af for Ax sufficiently small and positive and 
Ay = would be opposite to the sign of Af for Ax sufficiently small and 
negative and Ay = 0. Therefore for a minimum or maximum f x = ; 
and in like manner fy = 0. Considerations like these will serve to 
establish a criterion for distinguishing between maxima and minima 



PABTIAL DIFFERENTIATION; EXPLICIT 115 

analogous to the criterion furnished by f"( x ) in the ease °f one vari- 
able. For if f' x =fy = 0, then 

\f= \Q l fxx + 2 hkfx' y + kfyy) x=a + eh, y =b + eii 

by Taylor's Formula to two terms. Now if the second derivatives are 
continuous functions of (x, y) in the neighborhood of (a, b), each deriv- 
ative at (a -+- Oh, b -4- Oh) may be written as its value at (a, b) plus an 
infinitesimal. Hence 

Now the sign of Af for sufficiently small values of h, k must be the 
same as the sign of the first parenthesis provided that parenthesis does 
not vanish. Hence if the quantity 

fh*f" 4- 9 hhf" 4- TcH"\ > ° f ° r eV6iy &' ^ a minimUm 

1 ^ -T - nu Jxy -r n>JyyXa,V < q for every ^ ^ a maximum< 

As the derivatives are taken at the point (a, 5), they have certain constant 
values, say A, B, C. The question of distinguishing between minima and maxima 
therefore reduces to the discussion of the possible signs of a quadratic form 
Ah 2 + 2 Bhk + Ck 2 for different values of h and k. The examples 

h* + k 2 , -h 2 -k 2 , h 2 -k 2 , ±(h-k) 2 

show that a quadratic form may be : either 1°, positive for every (h, k) except (0, 0) ; 
or 2°, negative for every (h, k) except (0, 0) ; or 3°, positive for some values (h, k) 
and negative for others and zero for others ; or finally 4°, zero for values other than 
(0, 0), but either never negative or never positive. Moreover, the four possibilities 
here mentioned are the only cases conceivable except 5°, that A = B = C = and 
the form always is 0. In the first case the form is called a definite positive form, in 
the second a definite negative form, in the third an indefinite form, and in the fourth 
and fifth a singular form. The first case assures a minimum, the second a maxi- 
mum, the third neither a minimum nor a maximum (sometimes called a minimax) ; 
but the case of a. singular form leaves the question entirely undecided just as the 
condition f" (x) = did. 

The conditions which distinguish between the different possibilities may be ex- 
pressed in terms of the coefficients A, B, C. 

l°pos. def., B 2 <AC, A, C> ; 3° indef ., B 2 > AC ; 
2° neg. def., B 2 < AC, A, C < ; 4° sing., B 2 = AC. 

The conditions for distinguishing between maxima and minima are : 



fx = °\ f ,n f „ f „ \fZo f'yy > ° minimum ; 
/; = j Jxy JxxJ ^ l-CC < ° maximum ; 

fxy > fxxfyy, minimax ; f£ = f^f^ ( ?) . 



(34) 



It may be noted that in applying these conditions to the case of a definite form it 
is sufficient to .show that either/^ or/^ is positive or negative because they neces- 
sarily have the same sign. 



116 DIFFEBENTIAL CALCULUS 

EXERCISES 

1. Write at length, without symbolic shortening, the expansion of f(x, y) by- 
Taylor's Formula to and including the terms of the third order in x — a, y — b. 
Write the formula also with the terms of the third order as the remainder. 

2. Write by analogy the proper form of Taylor's Formula for/(x, ?/, z) and 
prove it. Indicate the result for any number of variables. 

3. Obtain the quadratic and lower terms in the development 

(a) of xy 2 + sin xy at (1, \ ir) and (/3) of tan- 1 (y/x) at (1, 1). 

4. A rectangular parallelepiped with one vertex at the origin and three faces 
in the coordinate planes has the opposite vertex upon the ellipsoid 

x 2 /a 2 + y 2 /b 2 + z 2 /c 2 =l. 

Find the maximum volume. 

5. Find the point within a triangle such that the sum of the squares of its 
distances to the vertices shall be a minimum. Note that the point is the intersec- 
tion of the medians. Is it obvious that a minimum and not a maximum is present ? 

6. A floating anchorage is to be made with a cylindrical body and equal coni- 
cal ends. Find the dimensions that make the surface least for a given volume. 

7. A cylindrical tent has a conical roof. Find the best dimensions. 

8. Apply the test by second derivatives to the problem in the text and to any 
of Exs. 4-7. Discuss for maxima or minima the following functions : 

(a) x 2 y + xy 2 — x, (/3) x s + y s — x 2 y' 2 - \ (x 2 + y 2 ), 

(y) x 2 + 'y 2 + x + y, (5) i y 3 - xy 2 + x 2 y - x, 

( e ) x 3 + y 3 — 9 xy + 27, (f ) x 4 + ?/ 4 — 2 x 2 + 4 xy — 2 y 2 . 

9. State the conditions on the first derivatives for a maximum or minimum of 
function of three or any number of variables. Prove in the case of three variables. 

10. A wall tent with rectangular body and gable roof is to be so constructed as 
to use the least amount of tenting for a given volume. Find the dimensions. 

11. Given any number of masses m v m 2 , • • •, m n situated at (x x , y^, (x 2 , 2/ 2 ), • • •, 
(x n , y n ). Show that the point about which their moment of inertia is least is their 
center of gravity. If the points were (x v y v z x ), • • • in space, what point would 
make Smr 2 a minimum ? 

12. A test for maximum or minimum analogous to that of Ex. 27, p. 10, may 
be given for a function /(x, y) of two variables, namely : If a function is positive 
all over a region and vanishes upon the contour of the region, it must have a max- 
imum within the region at the point for which f' x = f' y = 0. If a function is finite 
all over a region and becomes infinite over the contour of the region, it must have 
a minimum within the region at the point for which f' x = f = 0. These tests are 
subject to the proviso that/^ =fy = Q has only a single solution. Comment on the 
test and apply it to exercises above. 

13. If a, 6, c, r are the sides of a given triangle and the radius of the inscribed 
circle, the pyramid of altitude h constructed on the triangl e as base will have its 
maximum surface when the surface is | (a + b + c) Vr 2 + h 2 . 






CHAPTER V 

PARTIAL DIFFERENTIATION; IMPLICIT FUNCTIONS 

56. The simplest case ; F(x, y) = 0. The total differential 

dF = F' x dx + F'ydy = dO = 

• t dy Fx d x Fy /-ix 

indicates -£ = — -£-, — = — -* (1) 

<fe f; ^ f; v > 

as the derivative of y by oc, or of x by ?/, where ?/ is defined as a function 
of x, or x as a function of y, by the relation F(x, y) = ; and this method 
of obtaining a derivative of an implicit function without solving expli- 
citly for the function has probably been familiar long before the notion 
of a partial derivative was obtained. The relation F(x, y) = is pictured 
as a curve, and the function y = <f>(x) 3 which would be obtained by solu- 
tion, is considered as multiple valued or as restricted to some definite 
portion or branch of the curve F(x, y) = 0. If the results (1) are to 
be applied to find the derivative at some point 
(x Q , y ) of the curve F(x, y) = 0, it is necessary 
that at that point the denominator F' y or F' x should 
not vanish. 

These pictorial and somewhat vague notions 
may be stated precisely as a theorem susceptible 
of proof, namely : Let x Q be any real value of x 
such that 1°, the equation F(x , y) = has a real solution y ; and 2°, the 
function F(x, y) regarded as a function of two independent variables 
(x, y) is continuous and has continuous first partial derivatives F x , F' y in 
the neighborhood of (.r , y ) ; and 3°, the derivative F^(x , y )=£0 does 
not vanish for (x Q , y ) ; then F(x, y) = may be solved (theoretically) 
as y = 4> (x) in the vicinity of x = x and in such a manner that 
y = <£(.r ), that <f>(x) is continuous in x, and that 4>(x) has a derivative 
4>'( x ) — — F'x/Fy I and the solution is unique. This is the fundamental 
theorem on implicit functions for the simple case, and the proof follows. 

By the conditions on F x , F' y , the Theorem of the Mean is applicable. Hence 

F(x, y) -F(x , ?/ ) W>, y) = (hF x + kF^ + 0ht yn + ,,. (2) 

Furthermore, in any square |ft|<5, \k}<5 surrounding (x . y ) and sufficiently 
small, the continuity of F' insures I F' \ < M and the continuitv of F' taken with 

117 




X 



118 



DIFFEEEXTIAL CALCULUS 



the fact that F' (x , y ) ^ insures |F^|>m. Consider the range of x as further 

restricted to values such that \x — x | < md/M if m < M. Now consider the value 

of F(x, y) for any x in the permissible interval 

and for y = y + 8 or y = y — 5. As | kF' y | > md 

but | (x — x ) F' x | < md, it follows from (2) that 

F(x, y + 8) has the sign of SF y and F(x, y — 8) 

has the sign of — 8F' y ; and as the sign of F' y does 

not change, F(x, y + 5) and F{x, y — 8) have 

opposite signs. Hence by Ex. 10, p. 45, there is 

one and only one value of y between y — 8 and 

y + 5 such that F(x, y) = 0. Thus for each x in 

the interval there is one and only one y such 

that F(x, y) = 0. The equation F(x, y) = has a 

unique solution near (x , y ). Let y = <p(x) denote the solution. The solution is 

continuous at x = x because [ y — y 1 < 8. If (x, y) are restricted to values y = <j> (x) 

such that F(x, y) = 0, equation (2) gives at once 



Y 


X 


A/_LS 


8 




/ 


• Co ■ 


8 . 






- V 5 








#0 ° 





28 


m 
M 


X 



2/ 



Ay 
Ax 



F„{x + Oh, y + Ok) 

F' y tx + 0h,y + 0k) 



dy 

dx 



* X \ X Q •■ 



FifrwVo) 



As F' x , F' y are continuous and F' y ^ 0, the fraction k/h approaches a limit and the 
derivative / (x o ) exists and is given by (1). The same reasoning would apply to 
any point x in the interval. The theorem is completely proved. It may be added 
that the expression for 0'(x) is such as to show that 0'(x) itself is continuous. 

The values of higher derivatives of implicit functions are obtainable 
by successive total differentiation as 



K x + 2 F xy!/ > + F yy y^ + F' v y" = 0, 



(3) 



etc. It is noteworthy that these successive equations may be solved for 
the derivative of highest order by dividing by F y which has been assumed 
not to vanish. The question of whether the function y = <j> (x) denned 
implicitly by F(x, y) = has derivatives of order higher than the first 
may be seen by these equations to depend on whether F(x, y) has 
higher partial derivatives which are continuous in (x, y). 

57. To find the maxima and minima of y = <f> (x), that is, to find the 
points where the tangent to F(x, y) ' = ;0 is parallel to the ,T-axis, observe 
that at such points y' = 0. Equations (3) give 



K=0, F: x + F'y"=0. 



(4) 



Hence always under the assumption that F y ^ 0, there are maxima at 
the intersections of F = and F x = if F xx and F y have the same sign, 
and minima at the intersections for which F xx and F y have opposite signs; 
the case F xx = still remains undecided. 



PAETIAL DIFFERENTIATION; IMPLICIT 119 

For example if F(x, y) — x 3 + y 3 — 3 axy = 0, the derivatives are 

dy x 2 — ay 



S( x 2-ay) + S(y 2 -ax)y' = 0, 

6 z - 6 ay' + 6 yy' 2 + 3 (y 2 - ax) y" = 0, 



dx y 2 — ax 

d 2 y 2 a 3 xy 



dx 2 (y 2 — ax) 3 

To find the maxima or minima of y as a function of x, solve 

F x = = x 2 - ay, F = = X s + y 3 - 3 axy, F y ^ 0. 

The real solutions of F x = and F = are (0, 0) and (jy^2 a, J 7 ! a) of which the 
first must be discarded because F^(0, 0) = 0. At (^2 a, ^4 a) the derivatives 
F' and iv,, are positive ; and the point is a maximum. The curve F = is the 
folium of Descartes. 

The role of the variables x and y may be interchanged if F' x =£ and 
the equation F(x, y) = may be solved for x = ^(j/), the functions <£ 
and .iff being inverse. In this way the vertical tangents to the curve 
F = may be discussed. For the points of F = at which both F' x = 
and Fy = 0, the equation cannot be solved in the sense here defined. 
Such points are called singular points of the curve. The questions of 
the singular points of F = and of maxima, minima, or minimax (§ 55) 
of the surface z = F(x, y) are related. For if F x — F' y =p 0, the surface 
has a tangent plane parallel to z = 0, and if the condition z = F == is 
also satisfied, the surface is tangent to the xy-plane. Now if z = F(x, y) 
has a maximum or minimum at its point of tangency with z = 0, the 
surface lies entirely on one side of the plane and the point of tangency 
is an isolated point of F(x, y) = ; whereas if the surface has a mini- 
max it cuts through the plane z = and the point of tangency is not 
an isolated, point of F(x, y) = 0. The shape of the curve F = in the 
neighborhood of a singular point is discussed by developing F(x, y) 
about that point by Taylor's Formula. 

For example, consider the curve F(x, y) = x z + y 3 — x 2 y 2 — \ {x 2 -f y 2 ) = and 
the surface z = F(x, y). The common real solutions of 

F x = 3 x 2 -2xy 2 -x = 0, F' y = 3 y 2 - 2 x 2 y - y = 0, F(x, y) = 

are the singular points. The real solutions of F x = 0, Fy = are (0, 0), (1, 1), 
(!, |) and of these the first two satisfy F(a;, y) = but the last does not. The 
singular points of the curve are therefore (0, 0) and (1, 1). The test (34) of § 55 
shows that (0, 0) is a maximum for z — F(x,y) and hence an isolated point of 
F(x. y) = 0. The test also shows that (1, 1) is a minimax. To discuss the curve 
F(x, y) = near (1, 1) apply Taylor's Formula. 

= F(x, y) = \ (3 h 2 - 8 hk + 3 k 2 ) + i (6 A 3 - 12 h 2 k - 12 M- 2 + 6 fe 3 ) + remainder 
= \ (3 cos' 2 0—8 sin cos + 3 sin" 2 0) 

+ r (cos 3 — 2 cos 2 sin — 2 cos sin 2 -f sin 3 0) + • • • . 



120 DIFFERENTIAL CALCULUS 

if polar coordinates h = r cos 0, k = r sin be introduced at (1, 1) and r 2 be can- 
celed. Now for very small values of r, the equation can be satisfied only when 
the first parenthesis is very small. Hence the solutions of 

3 — 4 sin 2 = 0, sin 2 = f , or = 24° 17-1-', 65° 42|', 

and + 7T, are the directions of the tangents to F(x, y)= 0. The equation F = is 

= (li — 2 sin 2 0) + r (cos + sin 0) (1 — 11 sin 2 0) 

if only the first two terms are kept, and this will serve to sketch F(x, y) = for 
very small values of r, that is, for <p very near to the tangent directions. 

58. It is important to obtain conditions for the maximum or minimum 
of a function z = f(x, y) where the variables x, y are connected by a 
relation F(x, y) = so that z really becomes a function of x alone or y 
alone. For it is not always possible, and frequently it is inconvenient, 
to solve F(x, y) = for either variable and thus eliminate that variable 
from z = f(x, y) by substitution. When the variables x, y in z = f(x, y) 
are thus connected, the minimum or maximum is called a constrained 
minimum or maximum ; when there is no equation F(x, y) = between 
them the minimum or maximum is called free if any designation is 
needed. # The conditions are obtained by differentiating z =f(x, y) 
and F(x, y)=0 totally with respect to x. Thus 

dz_df ,dfdy__ () (]0_cF ^£(J]£_ () 

dx ex By dx dx ex cy dx 



dO 


dF 


dx 


ex 


d 2 z 
dx 2 


so, 



where the first equation arises from the two above by eliminating dy/dx 
and the second is added to insure a minimum or maximum, are the con- 
ditions desired. Note that all singular points of F(x, y) — satisfy the 
first condition identically, but that the process by means of which it 
was obtained excludes such points, and that the rule cannot be expected 
to apply to them. 

Another method of treating the problem of constrained maxima and 
minima is to introduce a multiplier and form the function 

z = &(x, y) =f(x, y) + XF(x, y), X a multiplier. (6) 

Now if this function z is to have a free maximum or minimum, then 

^ = f x + xf: c = o, $; = f y + xf;, = o. (7) 

These two equations taken with F = constitute a set of three from 
which the three values x, y, X may be obtained by solution. Note that 

* The adjective "relative" is sometimes used for constrained, and "absolute" for 
free ; but the term " absolute " is best kept for the greatest of the maxima or least of 
the minima, and the term " relative " for the other maxima and minima. 



PAETIAL DIFFERENTIATION; IMPLICIT 121 

X cannot be obtained from (7) if both F' x and F" y vanish ; and hence this 
method also rejects the singular points. That this method really deter- 
mines the constrained maxima and minima of f(x, y) subject to the 
constraint F(x, y) = is seen from the fact that if X be eliminated from 
(7) the condition f x F' y — f' y F' x = of (5) is obtained. The new method 
is therefore identical with the former, and its introduction is more a 
matter of convenience than necessity. It is possible to show directly 
that the new method gives the constrained maxima and minima. For 
the conditions (7) are those of a free extreme for the function <3> (x, y) 
which depends on two independent variables (x, y). Now if the equa- 
tions (7) be solved for (x, y), it appears that the position of the maximum 
or minimum will be expressed in terms of A. as a parameter and that 
consequently the point (x(k), y(fy) cannot in general lie on the curve 
F(x, y) = ; but if A be so determined that the point shall lie on this 
curve, the function 3>(a?, y) has a free extreme at a point for which 
F = and hence in particular must have a constrained extreme for the 
particular values for which F(x, y) = 0. In speaking of (7) as the con- 
ditions for an extreme, the conditions which should be imposed on 
the second derivative have been disregarded. 

For example, suppose the maximum radius vector from the origin to the folium 
of Descartes were desired. The problem is to render /(x, y) = x' 2 + y' 2 maximum 
subject to the condition _F(x, y) = x 6 + y s — 3 axy = 0. Hence 

2 x + 3 X (x 2 - ay) = 0, 2 y + 3 X (if- - ax) = 0, x s + y 3 - 3 axy = 

or 2 x • 3 (y 2 - ax) -2y-S(x 2 - ay) = 0, x s + y 3 - 3 axy = 

are the conditions in the two cases. These equations may be solved for (0, 0), 
(l|a, l|a), and some imaginary values. The value (0, 0) is singular and X cannot 
be determined, but the point is evidently a minimum of x 2 + y' 2 by inspection. The 
point (1^ a, li a) gives X = — 11 a. That the point is a (relative constrained) maxi- 
mum of x' 2 + y 2 is also seen by inspection. There is no need to examine d 2 f. In 
most practical problems the examination of the conditions of the second order 
may be waived. This example is one which may be treated in polar coordinates 
by the ordinary methods ; but it is noteworthy that if it could not be treated that 
way, the method of solution by eliminating one of the variables by solving the 
cubic F(x, y) = would be unavailable and the methods of constrained maxima 
would be required. 

EXERCISES 

1. By total differentiation and division obtain dy/dx in these cases. Do not 
substitute in (1), but use the method by which it was derived. 

(a) ax 2 + 2 bxy + cy 2 —1 = 0, (/3) z 4 + y 4 = 4 a' 2 xy, (y) (cosx)i> — (sin y) x = 0, 
(5 ) (x 2 + y 2 ) 2 = a 2 (x 2 - y 2 ), (e) & + &> = 2 xy, (f) x~ - 2 y- 2 = tan-i xy. 

2. Obtain the second derivative d 2 y/dx 2 in Ex. 1 (a), (/3), (e), (£) by differen- 
tiating the value of dy/dx obtained above. Compare with use of (3). 



122 DIFFEKENTIAL CALCULUS 

rTh/ Y' 2 F" — 2'F'F'F" 4- F' 2 F" 

3 Prove — = — y x % < *y^ * vv m 

dx 2 F /s 

y 

4. Find the radius of curvature of these curves : 

(a) xs + y~s = at , R = 3 (axy)s, (p) x^ + y? = ah, R = 2 V(x + yf/a, 
(7) & 2 x 2 + arij 2 = a 2 6 2 , (5) x?/ 2 = a 2 (a - x), (e) (ax) 2 + (&y)* = 1. 

5. Find 2/', 7/" y'" in case x 3 + ?/ 3 — 3 ax?/ = 0. 

6. Extend equations (3) to obtain y'" and reduce by Ex. 3. 

7. Find tangents parallel to the x-axis for (x 2 + y 2 ) 2 = 2 a 2 (x 2 — ?/ 2 ). 

8. Find tangents parallel to the y-axis for (x 2 + y 2 + ax) 2 = a 2 (x 2 + y 2 ). 

9 . If b 2 <ac in ax 2 + 2 bxy + cy 2 + /x + gy + A = 0, circumscribe about the 
curve a rectangle parallel to the axes. Check algebraically. 

10. Sketch x 3 + y s = xHj 2 + \ (x 2 + y 2 ) near the singular point (1, 1). 

11. Find the singular points and discuss the curves near them : 

(a) x 3 + y* = 3 axy, * m {p) (x 2 + y 2 ) 2 = 2 a 2 (x 2 - y% 

(7) x 4 + 2/ 4 = 2 (x - ?/) 2 , (5) y 5 + 2x# 2 = x 2 + yK 

12. Make these functions maxima or minima subject to the given conditions. 
Discuss the work both with and without a multiplier : 

, , a b , , . ■ sinx u 

(a) -H , atanx + b tan 2/ = c. -4ns. = -• 

it cos x 1; cos ?/ sin ?/ v 

(j3) x 2 + j/ 2 , ax 2 + 2 5x?/ + ci/ 2 =/. Find axes of conic. 

(7) Find the shortest distance from a point to a line (in a plane). 

13. Write the second and third total differentials of F(x, y) = and compare 
with (3) and Ex. 5. Try this method of calculating in Ex. 2. 

14. Show that F' x dx + F' y dy = does and should give the tangent line to 
F(x, y) = at the points (x, y) if dx = £ — x and dy = 77 — y, where £, rj are the 
coordinates of points other than (x, y) on the tangent line. Why is the equation 
inapplicable at singular points of the curve ? 

59. More general cases of implicit functions. The problem of 
implicit functions may be generalized in two ways. In the first place 
a greater number of variables may occur in the function, as 

^(^ y, 2) = 0, F(x, y, z, ■ • • , u) = ; 

and the question may be to solve the equation for one of the variables 
in terms of the others and to determine the partial derivatives of the 
chosen dependent variable. In the second place there may be several 
equations connecting the variables and it may be required to solve the 
equations for some of the variables in terms of the others and to 
determine the partial derivatives of the chosen dependent variables 



PARTIAL DIFFERENTIATION; IMPLICIT 123 

with respect to the independent variables. In both cases the formal 
differentiation and attempted formal solution of the equations for the 
derivatives will indicate the results and the theorem under which, the 
solution is proper. 

Consider the case F(x, y, z) = and form the differential. 

dF(x, y, z) = F'Jx + F' y dy + F' z dz = 0. (8) 

If z is to be the dependent variable, the partial derivative of z by x is 
found by setting dy = so that y is constant, Thus 

cz (dz\ F;. cz (dz\ F' y 

are obtained by ordinary division after setting dy = and dx = re- 
spectively. If this division is to be legitimate, F' z must not vanish at 
the point considered. The immediate suggestion is the theorem : If, 
when real values (x Q , y Q ) are chosen and a real value z Q is obtained 
from F(z, x Q , y ) = by solution, the function. F(x, y, z) regarded as 
a function of three independent variables (x, y, z) is continuous at 
and near (x , y , z Q ) and has continuous first partial derivatives and 
F z (x , y , z Q ) ^ 0, then F(x, y, z) = may be solved uniquely for 
z = <f> (x, y) and <£ (x, y) will be continuous and have partial derivatives 
(9) for values of (x, y) sufficiently near to (x Qj y ). 

The theorem is again proved by the Law of the Mean, and in a similar manner. 

F(x, y, z) - F(x , y , z ) = F(x, y, z) = QiF' x + l:F^ + iF e % +w ,y + Wfab + w . 

As I\'. Fy, F' z are continuous and F' z (x Q , y , z ) ^ 0. it is possible to take 5 so 
small that, when | h \ < <5, ( k \ < 5, \ 1 1 < 5, the derivative \F z \> m and | F' x | < /*, I F' y | < /*. 
Nov it is desired so to restrict h, k that ± dF' z shall determine the sign of the 
parenthesis. Let 

1^ — Zol<i m5 /^ \V— y \<h md /f*i then \hFx + kFy\<m8 

and the signs of the parenthesis for (x, y, z + 5) and (x, y, z — 5) will be opposite 
since j F' z ] > m. Hence if (x, y) be held fixed, there is one and only one value of z 
for which the parenthesis vanishes between z + 5 and z — 5. Thus z is defined as a 
single valued function of (x, y) for sufficiently small values of h = x — x Q , k = y — y . 

Also l - F *( x o +9h 'Vo + 0k - z o+ 0l ) i l = F v(--') 

h FZ(z o +0h,y o +0k,z o + 0l)' fc F' s {...) 

when k and h respectively are assigned the values 0. The limits exist when h = or 
k = 0. But in the first case I = Az = A x z is the increment of z when x alone varies, 
and in the second case I = Az =A y z. The limits are therefore the desired partial 
derivatives of z by x and y. The proof for any number of variables would be 
similar. 



124 DIFFEEENTIAL CALCULUS 

If none of the derivatives F' x ,F' y , F' z vanish, the equation F(x, y,z) = 
may be solved for any one of the variables, and formulas like (9) will 
express the partial derivatives. It then appears that 

dz\ fdx\ dz dx Fl Fi 

1, (10) 



dx) y \dz) y dx dz F' z F' x 

/dz\ /dx\ (dy\ _ dz dx dy _ 

\dx) y \dy) z \dz) x dx dy dz ^ ' 

in like manner. The first equation is in this case identical with (4) 
of § 2 because if y is constant the relation F(x, y, z) = reduces to 
G (x, z) = 0. The second equation is new. By virtue of (10) and simi- 
lar relations, the derivatives in (11) may be inverted and transformed 
to the right side of the equation. As it is assumed in thermodynamics 
that the pressure, volume, and temperature of a given simple substance 
are connected by an equation F(p, v, T) = 0, called the characteristic 
equation of the substance, a relation between different thermodynamic 
magnitudes is furnished by (11). 

60. In the next place suppose there are two equations 

F(x, y, u, v) = 0, G (x, y, u, v) = (12) 

between four variables. Let each equation be differentiated. 

dF = = F'JLx + Fydy + F' u du + F' v dv t 

dG = 0= G x dx + Gydy + G' u du + U' v dv. (13) 

If it be desired to consider u, v as the dependent variables and x, y as 
independent, it would be natural to solve these equations for the differ- 
entials die and dv in terms of dx and dy ; for example, 

7 (F' X G' V - F' v G x )dx + {F' y G' v - F' v G' v )dy 

F'G' -F'G' ' V- 6 ' 

U V V u 

The differential dv would have a different numerator but the same de- 
nominator. The solution requires F' u G' v — F' v G' u =£ 0. This suggests the 
desired theorem : If (w , v Q ) are solutions of F = 0, G = corresponding 
to (x , y ) and if F' U G' V — F' V G' U does not vanish for the values (x , y , u Q , v Q ), 
the equations F = 0, G = may be solved for u = cf>(x, y), v = if/(x, y) 
and the solution is unique and valid for (x, y) sufficiently near (x Q , y ) 
— it being assumed that F and G regarded as functions in four variables 
are continuous and have continuous first partial derivatives at and near 
(x , y , tc Q , v ) ; moreover, the total differentials du, dv are given by (13') 
and a similar equation. 



PAETIAL DIFFERENTIATION; IMPLICIT 125 

The proof of this theorem may be deferred (§ 64). Some observations 
should be made. The equations (13) may be solved for any two vari- 
ables in terms of the other two. The partial derivatives 

8" (■*'> //) du (-'', '0 dx ( u , v) ^{u, y) 

~ > n J n ' ~ (14) 

CX Cx CU GU 

of u by x or of x by u will naturally depend on whether the solution 
for u is in terms of (x, y) or of (x, v), and the solution for x is in (u } v) 
or (u, y). Moreover, it must not be assumed that du/dx and dx/du are 
reciprocals no matter which meaning is attached to each. In obtaining 
relations between the derivatives analogous to (10), (11), the values of 
the derivatives in terms of the derivatives of F and G may be found or 
the equations (12) may first be considered as solved. 

Thus if 



Then 



u = 0(r, y), 


du = <p x dx + 4>' y dy, 


v = yp (x, y), 


dv = ^ x dx + \f/' y dy. 


$x¥y - 4>'yVx 


- K du + 0> 

dy = — — t~t 

<$>x^y ~ <t>ytx 


CX _ Vy 


dx - - +'v c 


cu ;^_ ;^' 


dv ^ - jfe 


cu dx 
dx du 


dv dx 
■ — — — 1 » 

dx dv 



and — = , — = , etc. 



Hence 1 = 1 , (15) 

dx du dx dv 

as may be seen by direct substitution. Here u, v are expressed in terms of x, y for 
the derivatives u x , v x ; and x, y are considered as expressed in terms of u, v for the 



61. The questions of free or constrained maxima and minima, at any 
rate in so far as the determination of the conditions of the first order is 
concerned, may now be treated. If F (x, y, z) = is given and the max- 
ima and minima of z as a function of (x, y) are wanted, 

F' x (x, y, z) = 0, F' w (x, y, z) = 0, F(x, y, *) = (16) 

are three equations which may be solved for x, y, z. If for any of these 
solutions the derivative F' z does not vanish, the surface z — <f> (x, y) has 
at that point a tangent plane parallel to z = and there is a maximum, 
minimum, or minimax. To distinguish between the possibilities further 
investigation must be made if necessary ; the details of such an investi- 
gation will not be outlined for the reason that special methods are 
usually available. The conditions for an extreme of u as a function of 
(x, y) defined implicitly by the equations (13') are seen to be 

f&-f;g^=o, f;g;-f;g;=o, f=o, g = o. (it) 

The four equations may be solved for x, y, u, v or merely for x, y. 



126 DIFFEBENTIAL CALCULUS 

Suppose that the maxima, minima, and minimax of u = f(x, y, z) sub- 
ject either to one equation F(x, y, z) = or two equations F(x, y, z) = 0, 
G (x, y, z) = of constraint are desired. Note that if only one equation 
of constraint is imposed, the function u = fix, y, z) becomes a function 
of two variables ; whereas if two equations are imposed, the function u 
really contains only one variable and the question of a minimax does 
not arise. The method of multiiiliers is again employed. Consider 

<P(x,y,z)=f+XF or <D=/-fAF+/*G (18) 

as the case may be. The conditions for a free extreme of <3> are 

K=A, K=°> *.' = <>. (19) 

These three equations may be solved for the coordinates x, y, z which 
will then be expressed as functions of A. or of X and jjl according to the 
case. If then X or X and fi be determined so that (x, y, z) satisfy F = 
or F = and G = 0, the constrained extremes of u =f(x, y, z) will be 
found except for the examination of the conditions of higher order. 

As a problem in constrained maxima and minima let the axes of the section of 
an ellipsoid by a plane through the origin be determined. Form the function 



<£ = x 2 + 



y 2 + Z 2 + x I— + |_ , + . ?L _ 1 ) + . p (ix + my + nz) 

\a 2 b 2 c 2 / 



by adding to x 2 + y 1 + 2 2 , which is to be made extreme, the equations of the ellipsoid 
and plane, which are the equations of constraint. Then apply (19). Hence 

x + \- + ^l = 0, 2/ + x| + ^m = 0, z + \4 + % = 
a 2 2 b 2 2 c 2 2 

taken with the equations of ellipsoid and plane will determine x, y, z, X, /x. If the 
equations are multiplied by x, y, z and reduced by the equations of plane and 
ellipsoid, the solution for X is X =— r 2 =— (x 2 + y' 2 + z 2 ). The three equations 
then become 

1 ula 2 1 umb 2 1 ixnc 2 . jn 

x — , y — > z — , with Ix + my + nz = 0. 

2r 2 -a 2 J 2r 2 -b 2 2 r 2 - c 2 

Pa 2 m 2 b 2 n 2 c 2 . „ ._ A . 

Hence — + — + — = determines r 2 . (20) 

r 2 — a 2 r 2 — b 2 r 2 — c" 

The two roots for r are the major and minor axes of the ellipse in which the plane 
cuts the ellipsoid. The substitution of x, y, z above in the ellipsoid determines 

al \ 2 ( bm V / en \ 2 x 2 y 2 z 2 , 

Now when (20) is solved for any particular root r and the value of \x is found by 
(21), the actual coordinates x, y, z of the extremities of the axes may be found. 



PARTIAL DIFFERENTIATION; IMPLICIT 127 

EXERCISES 

1. Obtain the partial derivatives of z by x and y directly from (8) and not by 
substitution in (9). Where does the solution fail ? 

r' 2 v 2 z 2 1 

(«) - + r « + - = J » {P) x + y + z = —, 

a- o 2 c 2 xyz 

(7) (* 2 + y 2 + z 2 ) 2 = a 2 * 2 + BV + cV, (5) xyz = c. 

2. Find the second derivatives in Ex. 1 (a), (/3), (5) by repeated differentiation. 

3. State and prove the theorem on the solution of F(x, y, z, u) = 0. 

4. Show that the product a p E T of the coefficient of expansion by the modulus 
of elasticity (§ 52) is equal to the rate of rise of pressure with the temperature if 
the volume is constant. 

5. Establish the proportion E s : E r = C p : C v (see § 52). 

a t* t?i \ a i cucxcycz cu ex 

6. If F(x, y, z, u) = 0, show = 1, = 1. 

ex cy cz cu ex cu 

7. Write the equations of tangent plane and normal line to F(x, y. z) = and 
find the tangent planes and normal lines to Ex. 1 (/3), (5) at x = 1, y — 1. 

8. Find, by using (13), the indicated derivatives on the assumption that either 
x, y or m, v are dependent and the other pair independent : 

(a) u 5 + t> 5 + x 5 - 3 y = 0, u 3 + v 3 + y* + 3 x = 0, u^ u' y , u£, v^ 
(j8) x + y + u + v = a, x 2 + y 2 + a 2 + w 2 = &, &J, m*, V »^J 

(7) Find d?/ in both cases if x, u are independent' variables. 

9. Prove — ^ + — ^ = if F{x, y, u, v) = 0, G (x, y, w, 0) = 0. 

dx ou ex cu 

10. Find du and the derivatives u^ u' y , u' z in case 

x 2 + 2/ 2 + z 2 = uv, xy = u 2 + v 2 + w 2 , xyz = uvw. 

11. If F(x, y, z) = 0, 6?(x, 2/, z) = define a curve, show that 

s - So = y-Vo _ z ~ z o 

(K g ;-< g v\ (kk-kkx (k g ;-k g *\ 

is the tangent line to the curve at (x , y , z ). Write the normal plane. 

12. Formulate the problem of implicit functions occurring in Ex. 11. 

13. Find the perpendicular distance from a point to a plane. 

14. The sum of three positive numbers is x + y + z = 2V, where 2V is given. 
Determine x, y, z so that the product xPyiz r shall be maximum if p, g, r are given. 

^d?lS. X : y : Z : N = p : q : r : (p + q + r). 

15. The sum of three positive numbers and the sum of their squares are both 
given. Make the product a maximum or minimum. 

16. The surface (x 2 + y 2 + z 2 ) 2 = ax 2 + by' 2 -{-cz 2 is cut by the plane lx + my+nz = 0. 

Find the maximum or minimum radius of the section. Ans. > = 0. 

A' r 1 - a 



128 DIFFERENTIAL CALCULUS 

17. In case F(x, y, w, v) = 0, G(x, y, it, v) = consider the differentials 

, dv dv dx , dx , , dy , dy , 

du = — ax -1 ay, dx = — du -\ dv, dy = — du + — du. 

dx dy du dv du dv 

Substitute in the first from the last two and obtain relations like (15) and Ex. 9. 

18. If f{x, y, z) is to be maximum or minimum subject to the constraint 
F(x, y, z) = 0, show that the conditions are that dx : dy : dz = : : are indeter- 
minate when their solution is attempted from 

f x dx + f y dy + f z dz = and F x dx + F' y dy + F' z dz = 0. 

From what geometrical considerations should. t this be obvious ? Discuss in connec- 
tion with the problem of inscribing the maximum rectangular parallelepiped in 
the ellipsoid. These equations, 

dx : dy : dz =f y F' z -f'Xj ■•f z K-f x K •■f'Xy -fyK = 0:0:0, 
may sometimes be used to advantage for such problems. 

19. Given the curve F(x, y, z) = 0, G(x, y, z) = 0. Discuss the conditions for 
the highest or lowest points, or more generally the points where the tangent is 
parallel to z = 0, by treating u =f(x, y, z) = z as a maximum or minimum sub- 
ject to the two constraining equations F = 0, G = 0. Show that the condition 
F' x G' y — F' y G' x which is thus obtained is equivalent to setting dz = in 

F x dx + Fydy + F' z dz = and G x dx + G y dy + G' z dz = 0. 

20. Find the highest and lowest points of these curves : 

(a) x 1 + y* = z 2 + 1, x + y + 2 z = 0, (/3) — + V — + — = 1, Ix + my + nz = 0. 

a 2 b 2 c 2 

21. Show that F x dx + F' y dy + F z dz = 0, with dx = £ — x, dy = r) — y, dz = { — z, 
is the tangent plane to the surface F(x, y, z) = at (x, y, z). Apply to Ex. 1. 

22. Given F(x, y, it, v) = 0, G(x, y, u, v) = 0. Obtain the equations 

dF dFdu dFdv_ dF dF du dF dv _ 

dx du dx dv dx dy du dy dv dy 

dG dGdu oGdv . dG dGdu dG dv . 

^ + + = 0, — + H — — = 0, 

dx du dx dv dx dy du dy dv dy 

and explain their significance as a sort of partial-total differentiation of F = 
and G = 0. Find u' x from them and compare with (13'). Write similar equations 
where x, y are considered as functions of (u, v). Hence prove, and compare with 
(15) and Ex. 9, 

du dy dvdy _ du dx dv dx _ 

dy du dy dv dy du dy dv 

23. Show that the differentiation with respect to x and y of the four equations 
under Ex. 22 leads to eight equations from which the eight derivatives 

d 2 u d 2 u d 2 u 

dx 2 ' dxdy ' dydx 

may be obtained. Show thus that formally u 



d 2 u 
dy 2 ' 


d 2 v 
dx 2 ' 


d 2 v 
dy 2 


n y Kj 


= <• 





PAETIAL DIFFERENTIATION; IMPLICIT 129 

62. Functional determinants or Jacobians. Let two functions 

u = <j>(x,y), v = +(x,y) (22) 

of two independent variables be given. The continuity of the functions 
and of their first derivatives is assumed throughout this discussion 
and will not be mentioned again. Suppose that there were a relation 
F(u, v) = or F(cf>, if/) = between the functions. Then 

f(*,^) = o, f:^ + fm = o, F^;-h^>; = o. (23) 

The last two equations arise on differentiating the first with respect to 
x and y. The elimination of F' u and F' v from these gives 



Vx¥v — <t>ifx 






= l^ = j(^)=0. (24) 

d(x,y) \x,y) 



The determinant is merely another way of writing the first expression ; 
the next form is the customary short way of writing the determinant 
and denotes that the elements of the determinant are the first deriva- 
tives of u and v with respect to x and y. This determinant is called the 
functional determinant or Jacobian of the functions u, v or <£, \f/ with 
respect to the variables x, y and is denoted by J. It is seen that : If 
there is a functional relation F(<f>, <^r) = between tivo functions, the 
Jacobian of the functions vanishes identically, that is, vanishes for all 
values of the variables (x, y) under consideration. 

Conversely, if the Jacobian vanishes identically over a two-dimensional 
region for (x, y), the functions are connected by cc functional relation. 
For, the functions u, v may be assumed not to reduce to mere constants 
and hence there may be assumed to be points for which at least one of 
the partial derivatives §' n 4>' y , \\r' x , \j/y does not vanish. Let ^ be the 
derivative which does not vanish at some particular point of the region. 
Then u = <f>(x, y) may be solved as x = ^(», y) in the vicinity of that 
point and the result may be substituted in v. 

ti . fa C1 ' dx _ dv 1 , , , , n , ni , 

But — ==-— — and — = — (4>^, ~ VM (24') 

dy cycu cy <j> x y ^ 3 ' ru rx ^ yJ 

by (11) and substitution. Thus dv/dy = J/<f>^ ; and if J = 0, then 
dv/dy — 0. This relation holds at least throughout the region for which 
4>' r =£ 0, and for points in this region dv/dy vanishes identically. Hence 
v does not depend on y but becomes a function of u alone. This es- 
tablishes the fact that v and u are functionally connected. 



130 



DIFFERENTIAL CALCULUS 



These considerations may be extended to other cases. Let 
u = <f> (x, y, z), v = xf/ (x, y,z), iv = x (x, y, z). 

If there is a functional relation F(u, v, w) = 0, differentiate it. 

f& x + f& + f&=; o, 

KV V + KVy + K-xi = o, 
f' u & + f^' z + f; vX : = o, 



& 


V* 


Xx 


<*>; 


*i 


Xy 


+: 


v. 


Xz 



0, 



or 



3 (<k t, x) = d ( u > v > u ') = J= $ 

d(x, y, z) d(x, y, z). 



(25) 



(26) 



The result is obtained by eliminating F' u , F 7 V , F' w from the three equations. 
The assumption is made, here as above, that F' u , F' v , F' w do not all vanish ; 
for if they did, the three equations would not imply J = 0. On the 
other hand their vanishing would imply that F did not contain u, v, w, 
— as it must if there is really a relation between them. And now con- 
versely it may be shown that v if J vanishes identically, there is a func- 
tional relation between u, v, w. Hence again the necessary and sufficient 
conditions that the three functions (25) be functionally connected is that 
their Jacobian vanish. 



The proof of the converse part is about as before. It may be assumed that at 
least one of the derivatives of u, v, w or 0, \j/, x by %, V, z does not vanish. Let 
(p' x ■£. be that derivative. Then u = </>(#, y, z) may be solved as x = w(it, y, z) 
and the result may be substituted in v and w as 

v = \p{x,y,z) = i/ (w, y, z), w = x (x, y, z) = x («, y, z). 

Next the Jacobian of v and w relative to y and z may be written as 



cv 


cw 




cy 


cy 




cv 


dw 




cz 


dz 





,dx 

x x — 
cy 


+ 


Xy 


, ex 

*•* 


+ 


Xz 



+ +a 



<t>x L \^z X z 






+■£ 



Xy <t> % 

Xz 0, 



+ Xa 



+ X X 



K ~<t> 


V Mx 




K Vy 


_ j 


+z Vz 


4 


>x 



As J vanishes identically, the Jacobian of v and w expressed as functions of y, z, 
and u vanishes. Hence by the case previously discussed there is a functional- rela- 
tion F(v, w) . = independent of y, z ; and as v, w now contain u, this relation may 
be considered as a functional relation between u, v, w. 

63. If in (22) the variables u, v be assigned constant values, the 
equations define two curves, and if u, v be assigned a series of such 
values, the equations (22) define a network of curves in some part of the 



PAETIAL DIFFEEEXTIATIOX : IMPLICIT 



131 



sc^-plane. If there is a functional relation u = F(v), that is, if the 
Jacobian vanishes identically, a constant value of v implies a constant 
value of u and hence the locus for which v is constant is also a locus 
for which u is constant ; the set of r-curves coincides with the set of 
^-curves and no true network is formed. This ^ 

case is uninteresting. Let it be assumed that 
the Jacobian does not vanish identically and 
even that it does not vanish for any point (x, y) 
of a certain region of the ^y-plane. The indi- 
cations of § 60 are that the equations (22) may 
then be solved for x, y in terms of u, v at any 
point of the region and that there is a pair of 
the curves through each point. It is then proper to consider (u, v) as 
the coordinates of the points in the region. To any point there corre- 
spond not only the rectangular coordinates (x, y) but also the curvi- 
linear coordinates (u, r). 

The equations connecting the rectangular and curvilinear coordinates 
may be taken in either of the two forms 



Y 


l\ 









X 



u=<f> (x, y), v = $ (x, y) or x = f(u, v), y = g (u, v), (22') 

each of which are the solutions of the other. The Jacobians 

2/ N 



\x, y) \u, v 



are reciprocal each to each ; and this rela- 
tion may be regarded as the analogy of 
the relation (4) of § 2 for the case of 
the function y = <f> (x) and the solution 
x = fiy) = <£ -1 (#) in the case of a single 
variable. The differential of are is 



= 1 
Y 



(27) 



(x+d v x, y+d v y) 
(u, v+dv) 

(x+dx,y+dy) 

(u + du, v+dv) 
v + dv 



ds 2 = dx 2 -f- dif = Edir -f 2 Fdudr + Gdi 




\cu) \cu 



F 



ex ex 

Cll cc 



cy cy 

cu cr 



fdxV 

s)+ 



&)■■ 



The differential of area included between two neighboring ^-curves and 
two neighboring r-curves may be written in the form 



d A = J[ - 1 - JL - I dudv = dudv -r- J 



u, vj \x, y 

These statements will now be proved in detail. 



(29) 



132 



DIFFEKENTIAL CALCULUS 



To prove (27) write out the Jacobians at length and reduce the result. 



\x, y! \u, »/ 



du dv 
dx dx 




dx 
du 


dy 
du 








du dv 

dy dy 




dx 

dv 


dy 
cv 




du dx dv dx 
dx du dx dv 


du dy dv dy 
dx du dx dv 




1 


dudx 
dy du 


dv dx 
dy dv 


du dy dv dy 
dy du dy dv 




1 



where the rule for multiplying determinants has been applied and the reduction 
has been made by (15), Ex. 9 above, and similar formulas. If the rule for multi- 
plying determinants is unfamiliar, the Jacobians may be written and multiplied 
without that notation and the reduction may be made by the same formulas as 
before. 

To establish the formula for the differential of arc it is only necessary to write 
the total differentials of dx and dy, to square and add, and then collect. To obtain 
the differential area between four adjacent curves consider the triangle determined 
by (u, u), (u + du, v), (it, 15 + dv), which is half that area, and double the result. 
The determinantal form of the area of a triangle is the best to use. 






dA = 2> 



d u x d u y 
d v x d v y 



— du 

du 


d -ldu 
du 




dx dy 
du du 


— dv 

dv 


— dv 

dv 




dx dy 

dv dv 



dudv. 



The subscripts on the differentials indicate which variable changes ; thus d u x, d u y 
are the coordinates of (u + du, v) relative to (it, 15). This method is easily extended 
to determine the analogous quantities in three dimensions or more. It may be 
noticed that the triangle does not look as if it were half the area (except for infin- 
itesimals of higher order) in the figure ; but see Ex. 12 below. 

It should be remarked that as the differential of area clA is usually 
considered positive when du and dv are positive, it is usually better to 
replace J in (29) by its absolute value. Instead of regarding (u, v) as 
curvilinear coordinates in the ^?/-plane, it is possible to plot them in 
their own itv-plane and thus to establish by (22') a transformation of 
the ccy-plane over onto the wv-plane, A small area in the ;r?/-plane then 
becomes a small area in the wv-plane. If J > 0, the transformation is 
called direct ; but if J < 0, the transformation is called perverted. The 
significance of the distinction can be made clear only when the ques- 
tion of the signs of areas has been treated. The transformation is called 
conformal when elements of arc in the neighborhood of a point in the 
ic?/-plane are proportional to the elements of arc in the neighborhood of 
the corresponding point in the wv-plane, that is, when 






ds 2 = dx 2 + dif = h (du 2 + dv 2 ) = kdi 



(30) 



PAETIAL DIFFERENTIATION; IMPLICIT 133 

For in this case any little triangle will be transformed into a little tri- 
angle similar to it, and hence angles will be unchanged by the transfor- 
mation. That the transformation be conformal requires that F = and 
E = G. It is not necessary that E = G = k be constants ; the ratio of 
similitude may be different for different points. 

64. There remains outstanding the proof that equations may be solved 
in the neighborhood of a point at which the Jacobian does not vanish. 
The fact was indicated in § 60 and used in § 63. 

Theorem. Let p equations in n -f- p variables be given, say, 

F x (x l9 x 2 , .-.,x n +p ) = 0, F 2 = 0, • • • , F p = 0. (31) 

Let the p functions be soluble for x lo , x^, • • • , x po when a particular set 
x (p+l)o , • • •, x (n+p)o of the other n variables are given. Let the functions 
and their first derivatives be continuous in all the n + p variables in the 
neighborhood of (x , # 2o , • ■ -, x (n + p)<) ). Let the Jacobian of the functions 
with respect to x 



(32) 



9 X (n+p)o 

fail to vanish for the particular set mentioned. Then the p equations 
may be solved for the p variables x v x 2 , ■ ■ -,x p , and the solutions will be 
continuous, unique, and differentiable with continuous first partial 
derivatives for all values of x p+1 , •••, x n+p sufficiently near to the 
values x (p+1)o , •••, x (n+p)o . 

Theorem. The necessary and sufficient condition that a functional 
relation exist between pj functions of p variables is that the Jacobian 
of the functions with respect to the variables shall vanish identically, 
that is, for all values of the variables. 

The proofs of these theorems will naturally be given by mathematical induction. 
Each of the theorems has been proved in the simplest cases and it remains only to 
show that the theorems are true for p functions in case they are for p — 1. Expand 
the determinant J. 



X \1 X 2> ' 


*J ^J35 


dx 1 


."-111 




\x v • • 




' dx x 


¥= 






cx p 


cx p 


X lo 



dx x cx 2 p cx p 



Eor the first theorem J ^ and hence at least one of the minors J x , • • • , J p must 
fail to vanish. Let that one be J x , which is the Jacobian of F 2 , • ■ • , F p with respect 
to x 2 , • • •, x p . By the assumption that the theorem holds for the case p — 1, these 
p — 1 equations may be solved for x 2 , • • •. x p in terms of the n + 1 variables x 1 , 



1U 



DIFFERENTIAL CALCULUS 



•Ep +1; 



x n +p, and the results may be substituted in F r It remains to show that 



dF 1 _dF\,dF 1 cx^ 



± r ^ = J/J 1 ^0. (32') 

x p with respect to x x are obtained from the equations 
dF n 



p cF p dx 2 






For the derivatives of x s 

resulting from the differentiation of , F 2 = 0, • • ■, -Fp = with respect to x v The 
derivative cX[/cx x is therefore merely J^/J^ , and hence dF 1 /dx 1 = J/J x and does 
not vanish. The equation therefore may be solved for x x in terms of x p + 1 , • • • , 

X n +pi 



and this result may be substituted in the solutions above found for x. 2 



Hence the equations have been solved for x 15 x 2 , • • •, x p in terms of x p +i, • • • , x w+ p 
and the theorem is proved. 

For the second theorem the procedure is analogous to that previously followed. 
If there is a relation F(u x , • •, u p ) = between the p functions 

«i = 1 (» 1 , • • •, Xp), •. • •, u p = 0p (»!*, • 5 • , Xp), 



differentiation with respect to x x , 
tives of F by u x , 



Xp gives p equations from which the deriva- 
u p may be eliminated and J\ T 



= becomes the con- 



dition desired. If conversely this Jacobian vanishes identically and it be assumed 
that one of the derivatives of Ui by x,-, say du x /Zx x ^ does not vanish, then the solution 
Xj = w(u 1 , x 2 , • • •, Xp) may be effected and the result may be substituted in it 2 , 
• • ., Up. The Jacobian of u 2 , • • ■, u p with respect to x 2 , • • •, x p will then turn out 
to be J -H- cu 1 /cx 1 and will vanish because J vanishes. Now, however, only p — 1 
functions are involved, and hence if the theorem is true for p — 1 functions it must 
be true for p functions. 

EXERCISES 

1. If u = ax + by + c and v = a'x + b'y + e' are functionally dependent, the 
lines u = and v = are parallel ; and conversely. 

2. Prove x + y + z, xy + yz + zx, x 2 + y 2 + z 2 functionally dependent. 

3. If w = ax + by + cz + a*, v = a'x + b'y + c'z + d\ 10 — a"x + b"y + c' -/ z + cZ" 
are functionally dependent, the planes w = 0, u = 0, w = are parallel to a line. 

cv cZ.F 5JP 

4. In what senses are — and i// of (24') and — ! and — 1 of (32') partial or total 

dy dx 1 dx x 

derivatives ? Are not the two sets completely analogous ? 



Yy Xij 
Vz Xz 

tute in u = 0, and prove du/dx = , 



5. Given (25), suppose 



^ 0. Solve v = \p and w = x f or 2/ an( i 2, substi- 

6. If u — u (x, ?/), v = v (x, ?/), and x = x (£, y),y = y (£, 17), prove 

State the extension to any number of variables. How may (27') be used to prove 
(27) ? Again state the extension to any number of variables. 



(27') 



PARTIAL DIFFERENTIATION; IMPLICIT 135 

7. Prove dV = J I — — - — ) dudvdw = dudvdw -f- J (— — - — ) is the element of 

\u, v, wj \x, ?/, zf 

volume in space with curvilinear coordinates u, v, w = consts. 

8. In what parts of the plane can u = x 2 + y 2 , v = xy not be used as curvi- 
linear coordinates ? Express ds 2 for these coordinates. 

9. Prove that 2u = x 2 — y 2 , v = xy is a conformal transformation. 

u v 

10. Prove that x = , y = is a conformal transformation. 

u' 1 + v 2 u 2 + o- 

11. Define conformal transformation in space. If the transformation 
x = an + bv + cw, y = a'u + h'v + c'w, z = a"u + h"x> + c"w 

is conformal, is it orthogonal ? See Ex. 10 (f), p. 100. 

12. Show that the areas of the triangles whose vertices are 

(u, u), (u + du, v), (u, v + dv) and (u + du, v + dv), (u + du, v), (w, u + du) 
are infinitesimals of the same order, as suggested in § 63. 

13. "Would the condition F = in (28) mean that the set of curves u = const. 
were perpendicular to the set v = const. ? 

14. Express E, F, 6? in (28) in terms of the derivatives of it, u by x, y. 

15. If x = 0(s, £), y = \p (s, £), z = x (s, t) are the parametric equations of a 
surface (from which s, t could be eliminated to obtain the equation between 
x, y, z), show 

ex \s, t / \s, t / cy 

65. Envelopes of curves and surfaces. Let the equation F(x, y,a) = 
be considered as representing a family of curves where the different 
curves of the family are obtained by assigning different values to the 
parameter a. Such families are illustrated by 

(x — af -f- if = 1 and ax -f- y/a = 1, (33) 

which are circles of unit radius centered on the sc-axis and lines which 
cut off the area \ a 2 from the first quadrant. As a changes, the circles 
remain always tangent to the two lines i/=±l and 
the point of tangency traces those lines. Again, as Y 
a changes, the lines (33) remain tangent to the hyper- 
bola xy = 7x, owing to the property of the hyperbola 
that a tangent forms a triangle of constant area with 
the asymptotes. The lines y = ± 1 are called the - 
envelope of the system of circles and the hyperbola 
xy — k the envelope of the set of lines. In general, if there is a curve 
to which the curves of a family F(x, y, a) = are tangent and if the 
pjoint of tangency describes that curve as a varies, the curve is called 



* = jlxi±\^j('h±\ andflnd * 




136 DIFFERENTIAL CALCULUS 

the envelope (or part of the envelope if there are several such curves) 
of the family F(x, y, a) = 0. Thus any curve may be regarded as the 
envelope of its tangents or as the envelope of its circles of curvature. 

To find the equations of the envelope note that by definition the 
enveloping curves of the family F(x, y,a) = are tangent to the envelope 
and that the point of tangency moves along the envelope as a varies. 
The equation of the envelope may therefore be written 

x = 4>(a), y = if/(a) with F(<f>, if/, a) = 0, (34) 

where the first equations express the dependence of the points on the 
envelope upon the parameter a and the last equation states that each 
point of the envelope lies also on some curve of the family F(x, y, a) = 0. 
Differentiate (34) with respect to a. Then 

F' x <f>\a) + FtfXa) + F' a ~ 0. (35) 

Now if the point of contact of the envelope with the curve F = is an 
ordinary point of that curve, the tangent to the curve is 

F x (x-x ) + F^y-y ) = 0', and F x cj>' + Ftf = 0, 

since the tangent direction dy : dx = if/' : <f>' along the envelope is by 
definition identical with that along the enveloping curve ; and if the 
point of contact is a singular point for the enveloping curve, F' x — F' y = 0. 
Hence in either case F' a = 0. 

Thus for points on the envelope the two equations 

F(x, y, a) = 0, Ffr, y, a) = (36) 

are satisfied and the equation of the envelope of the family F = may 
be found by solving (36) to find the parametric equations x-=<j>(a), 
y = if/ (a) of the envelope or by eliminating a between (36) to find the 
equation of the envelope in the form <I> (x, y) = 0. It should be remarked 
that the locus found by this process may contain other curves than the 
envelope. For instance if the curves of the family F = have singular 
points and if x = <f> (a), y = if/ (a) be the locus of the singular points 
as a varies, equations (34), (35) still hold and hence (36) also. The 
rule for finding the envelope therefore finds also the locus of singular 
points. Other extraneous factors may also be introduced in performing 
the elimination. It is therefore important to test graphically or analyt- 
ically the solution obtained by applying the rule. 

As a first example let the envelope of (x — a)' 2 + y" = 1 be found. 

F(x, y, a) = (x- a) 2 + y* - 1 = 0, F' a — — 2 (x - a) = 0. 

The elimination of a from these equations gives y 2 — 1 = and the solution 
for a gives x = a, y =± 1. The loci indicated as envelopes are y = ±1. It is 



PARTIAL DIFFERENTIATION ; IMPLICIT 137 

geometrically evident that these are really envelopes and not extraneous factors. 
But as a second example consider ax + y/a = 1. Here 

F(x, y, a) = ax + y/a -1 = 0, F' a = x - y/a 2 = 0. 

The solution is y = a/2, x = 1/2 a, which gives xy = \. This is the envelope ; it could 
not be a locus of singular points of F = as there are none. Suppose the elimina- 
tion of a be made by Sylvester's method as 

— y/a 2 + 0/a + x + 0a = — y x 

0/a 2 - y/a + + xa = - y Ox 

?//o: 2 — l/« + x + 0a = ' 2/-1 z0 

0/a 2 +y/a — 1 + xa = */ — 1 a; 



= 0; 



the reduction of the determinant gives x?/ (4 xy — 1) = as the eliminant, and con- 
tains not only the envelope 4xz/ = 1, but the factors x = and ?/ = which are 
obviously extraneous. 

As a third problem find the envelope of a line of which the length intercepted 
between the axes is constant. The necessary equations are 

- + V - = 1, a 2 + p 2 = K 2 , — da + — dp = 0, ada + j3dj3 = 0. 
a p a 2 p 2 

Two parameters or, p connected by a relation have been introduced; both equations 
have been differentiated totally with respect to the parameters ; and the problem 
is to eliminate a, /3, da, dp from the equations. In this case it is simpler to carry 
both parameters than to introduce the radicals which would be required if only 
one parameter were used. The elimination of da, dp from the last two equations 
gives x : y = a 3 : p s or y/x : ^/y = a : p. From this and the first equation, 

1 1 1 1 , , 2 f 2 9 

and hence xs + yz = Ks. 



X3 



Hxi + yi) /S yi(»f + yl) 



66. Consider two neighboring curves of F(x, y, a) = 0. Let (x , y ) 
be an ordinary point of a = a Q and (x Q + dx, y 4- <%) of a 4- da. Then 

F(x + dx, y + rfy, tf 4 da) - F(x Q , y , a ) 

= F^fcc + ^2/ 4- K^ = (37) 

holds except for infinitesimals of higher order. The distance from the 
point on a 4 da to the tangent to a at (x Q , y ) is 

* &+*■& = f Kd « = dn (38) 

±VfY+F? Vf; 2 4- F? 

except for infinitesimals of higher order. This distance is of the first 
order with da, and the normal derivative da/dn of § 48 is finite except 
when F' a = 0. The distance is of higher order than da, and da/dn is 
infinite or dn/da is zero when F' a = 0. It appears therefore that the 
envelope is the locus of points at which the distance between two neigh- 
boring curves is of higher order than da. This is also apparent geomet- 
rically from the fact that the distance from a point on a curve to the 



138 DIFFERENTIAL CALCULUS 

tangent to the curve at a neighboring point is of higher order (§ 36). 
Singular points have been ruled out because (38) becomes indetermi- 
nate. In general the locus of singular points is not tangent to the 
curves of the family and is not an envelope but an extraneous factor ; 
in exceptional cases this locus is an envelope. 

If two neighboring curves F(x, y, a) = 0, F(x, y, a + Aa) = inter- 
sect, their point of intersection satisfies both of the equations, and hence 
also the equation 

— [F(x, y,a + Aa) - F(x, y, a)] = F' a (x, y, a + OAa) = 0. 

If the limit be taken for Aa = 0, the limiting position of the intersec- 
tion satisfies F' a = and hence may lie on the envelope, and will lie on 
the envelope if the common point of intersection is remote from singular 
points of the curves F(x, y, a) = 0. This idea of an envelope as 'the 
limit of points in which neighboring curves of the family intersect is 
valuable. It is sometimes taken as the definition of the envelope. But, 
unless imaginary points of intersection are considered, it is an inade- 
quate definition ; for otherwise y = (x — a) s would have no envelope 
according to the definition (whereas y = is obviously an envelope) and 
a curve could not be regarded as the envelope of its osculating circles. 

Care must be used in applying the rule for finding an envelope. Otherwise not 
only may extraneous solutions be mistaken for the envelope, but the envelope may 
be missed entirely. Consider 

y — sin ax = or a — x _1 sin -1 y = 0, (39) 

where the second form is obtained by solution and contains a multiple valued 
function. These two families of curves are identical, and it is geometrically clear 
that they have an envelope, namely y = ±1. This is precisely what would be 
found on applying the rule to the first of (39) ; but if the rule be applied to the 
second of (39), it is seen that F / a = l, which does not vanish and hence indicates no 
envelope. The whole matter should be examined carefully in the light of implicit 
functions. 

Hence let F(x, y, a) = be a continuous single valued function of the three 
variables (x, y, a) and let its derivatives F x , F^, F' a exist and be continuous. Con- 
sider the behavior of the curves of the family near a point (x , y ) of the curve for 
a = a provided that (x , y ) is an ordinary (nonsingular) point of the curve and 
that the derivative F^(x Q , y , a ) does not vanish. As F' a ■£ and either F' x ^ 
or F' y yt for (x , y , a ), it is possible to surround (x , y ) with a region so small 
that F(x, y, a) = may be solved for a — f(x, y) which will be single valued and 
differentiable ; and the region may further be taken so small that F x or F' y remains 
different from throughout the region. Then through every point of the region 
there is one and only one curve a =f(x, y) and the curves have no singular points 
within the region. In particular no two curves of the family can be tangent to 
each other within the region. 



PAETIAL DIFFERENTIATION; IMPLICIT 139 

Furthermore, in such a region there is no envelope. For let any curve which 
traverses the region be x = (t), y = \f/(t). Then 

a(t) = f(<f>(t), *(*)), a'(t) = f x <t>'{t)+f' y f{t). 

Along any curve a = /(x, y) the equation f x dx + f' y dy = holds, and if x = 0(t), 
y = \p(t) he tangent to this curve, dy — dx = \p' : <p' and tx'{t) = or a = const. 
Hence the only curve which has at each point the direction of the curve of the 
family through that point is a curve which coincides throughout with some curve 
of the family and is tangent to no other member of the family. Hence there is no 
envelope. The result is that an envelope can be present only when F' a = or when 
F' x — F' y = 0, and this latter case has been seen to be included in the condition 
F' a = 0. If F(x, ?/, a) were not single valued but the branches were separable, the 
same conclusion would hold. Hence in case F(x, y, a) is not single valued the loci 
over which two or more values become inseparable must be added to those over 
which F^ = in order to insure that all the loci which may be envelopes are taken 
into account. 

67. The preceding considerations apply with so little change to other 
cases of envelopes that the facts will merely be stated without proof. 
Consider a family of surfaces F(x, y, z, a, ft) = depending on two 
parameters. The envelope may be defined by the property of tangency 
as in § 65 ; and the conditions for an envelope would be 

F(x,y,z,a,P) = 0, F^ = 0, F, = 0. (-40) 

These three equations may be solved to express the envelope as 

x = <f>(a,(3), y = x\,(a,P), z = x (a, 0) 

parametrically in terms of a, ft; or the two parameters may be elimi- 
nated and the envelope may be found as <£ (x, y, z) = 0. In any case 
extraneous loci may be introduced and the results of the work should 
therefore be tested, which generally may be done at sight. 

It is also possible to determine the distance from the tangent plane 
of one surface to the neighboring surfaces as 

F^F^ + F^dz = F'ja + FW = ^ 

and to define the envelope as the locus of points such that this distance 
is of higher order than \da\ + \d(3'. The equations (40) would then also 
follow. This definition would apply only to ordinary points of the sur- 
faces of the family, that is, to points for which not all the derivatives 
F x , Fyj F' z vanish. But as the elimination of a, ft from (40) would give 
an equation which included the loci of these singular points, there 
would be no danger of losing such loci in the rare instances where they, 
too, happened to be tangent to the surfaces of the family. 



140 DIFFEBENTIAL CALCULUS 

The application of implicit functions as in § 66 could also be made in this case 
and would show that no envelope could exist in regions where no singular points 
occurred and where either F' a or F^ failed to vanish. This work could be based 
either on the first definition involving tangency directly or on the second definition 
which involves tangency indirectly in the statements concerning infinitesimals of 
higher order. It may be added that if F(x, y, z, a, fi) = were not single valued, 
the surfaces over which two values of the function become inseparable should be 
added as possible envelopes. 

A family of surfaces F(x, y, z, a) = depending on a single param- 
eter may have an envelope, and the envelope is found from 

F(x, y, z, a) = 0, F^x, y, z, a) = (42) 

by the elimination of the single parameter. The details of the deduction 
of the rule will be omitted. If two neighboring surfaces intersect, the 
limiting position of the curve of intersection lies on the envelope and 
the envelope is the surface generated by this curve as a varies. The 
surfaces of the family touch the envelope not at a point merely but 
along these curves. The curves are called characteristics of the family. 
In the case where consecutive surfaces of the family do not intersect 
in a real curve it is necessary to fall back on the conception of imagi- 
naries or on the definition of an envelope in terms of tangency or 
infinitesimals; the characteristic curves are still the curves along 
which the surfaces of the family are in contact with the envelope and 
along which two consecutive surfaces of the family are distant from 
each other by an infinitesimal of higher order than da. 

A particular case of importance is the envelope of a plane which 
depends on one parameter. The equations (42) are then 

Ax + By + Cz + D = 0, A'x -f B'y + C'z + D' = 0, (43) 

where A, B, C, D are functions of the parameter and differentiation 
with respect to it is denoted by accents. The case where the plane 
moves parallel to itself or turns about a line may be excluded as trivial. 
As the intersection of two planes is a line, the characteristics of the 
system are straight lines, the envelope is a ruled surface, and a plane 
tangent to the surface at one point of the lines is tangent to the surface 
throughout the whole extent of the line. Cones and cylinders are exam- 
ples of this sort of surface. Another example is the surface enveloped 
by the osculating planes of a curve in space ; for the osculating plane 
depends on only one parameter. As the osculating plane (§ 41) may be 
regarded as passing through three consecutive points of the curve, two 
consecutive osculating planes may be considered as having two consecu- 
tive points of the curve in common and hence the characteristics are 



PARTIAL DIFFERENTIATION; IMPLICIT 141 

the tangent lines to the curve. Surfaces which are the envelopes of a 
plane which depends on a single parameter are called developable surfaces. 
A family of curves dependent on two parameters as 

F(x, y, z, a, £) = <), G (x, y, z, a, (3) = (44) 

is called a congruence of curves. The curves may have an envelope, that 
is, there may be a surface to which the curves are tangent and which 
may be regarded as the locus of their points of tangency. The envelope 
is obtained by eliminating a, /3 from the equations 

f = o, £ = o, f:g;-f;gz = o. (45) 

To see this, suppose that the third condition is not fulfilled. The equa- 
tions (44) may then be solved as a =f(x, y, z), f$='g(x, y, z). Reason- 
ing like that of § 66 now shows that there cannot possibly be an 
envelope in the region for which the solution is valid. It may therefore 
be inferred that the only possibilities for an envelope are contained in 
the equations (45). As various extraneous loci might be introduced in 
the elimination of a, (3 from (45) and as the solutions should therefore 
be tested individually, it is hardly necessary to examine the general 
question further. The envelope of a congruence of curves is called the 
focal surface of the congruence and the points of contact of the curves 
with the envelope are called the focal points on the curves. 

EXERCISES 

1. Find the envelopes of these families of curves. In each case test the answer 
or its individual factors and check the results by a sketch : 

(a) y = 2 ax + a 4 , (j3) y 2 = a(x — a), (y) y = ax + Jc/a, 

(8) a(y + a) 2 = x 3 , (e) y = a(x + a) 2 , (f) y 2 = a(x - a) 3 . 

2. Find the envelope of the ellipses x 2 /a 2 + y 2 /b 2 = 1 under the condition that 
(a) the sum of the axes is constant or (/3) the area is constant. 

3. Find the envelope of the circles whose center is on a given parabola and 
which pass through the vertex of the parabola. 

4. Circles pass through the origin and have their centers on x 2 — y 2 = c 2 . Find 
their envelope. Ans. A lemniscate. 

5 . Find the envelopes in these cases : 

(a) x + xya = sin- x &?/, (/3) x + a = vers- l y + V2 y — y 2 % 
(7) y+a = Vl — 1/x. 

6. Find the envelopes in these cases : 

(a) ax + py+ a[3z = 1, (/3) - + V - + 



a |8 1—a — 

W ~2 + £ + -2 = 1 with a ^ = kK 



a- 



7. Find the envelopes in Ex. 6 (a), (/3) if a = /3 or if a = — /3. 



142 DIFFERENTIAL CALCULUS 

8. Prove that the envelope of JP(x, y, z, a) — is tangent to the surface along 
the whole characteristic by showing that the normal to F(x, y, z, a) = and to the 
eliminant of F = 0, F' a — are the same, namely 

F' :F' :F' and F' + F' — : F' + F' ~ • F' + F' — , 

where a (x, y, z) is the function obtained by solving F' a = 0. Consider the problem 
also from the point of view of infinitesimals and the normal derivative. 

9. If there is a curve x = 0(a), y = \f/(a), z = x( a ) tangent to the curves of 
the family defined by F (x, y, z, a) = 0, G (x, y, z, a) = in space, then that curve 
is called the envelope of the family. Show, by the same reasoning as in § 65 for 
the case of the plane, that the four conditions F = 0, G = 0, F' a — 0, G' a — must 
be satisfied for an envelope ; and hence infer that ordinarily a family of curves in 
space dependent on a single parameter has no envelope. 

10. Show that the family F(x, y, z, a) = 0, i^(x, y, z, a) = of curves which 
are the characteristics of a family of surfaces has in general an envelope given by 
the three equations F = 0, Fa = 0, F^ a = 0. 

11. Derive the condition (45) for the envelope of a two-parametered family of 
curves from the idea of tangency, ag in the case of one parameter. 

12. Find the envelope of the normals to a plane curve y =f(x) and show that 
the envelope is the locus of the center of curvature. 

13., The locus of Ex. 12 is called the evolute of the curve y =/(x). In these cases 
find the evolute as an envelope : 

(a) y = x 2 , (j3) x = asmt, y = bcost, (y) 2xy = a 2 , 

(5) y 2 = 2mx, (e) x = a(6— sin#), y = a(l — cos 6), (f) y = coshx. 

14. Given a surface z =/(x, y). Construct the family of normal lines and find 
their envelope. 

15. If rays of light issuing from a point in a plane are reflected from a curve in 
the plane, the angle of reflection being equal to the angle of incidence, the envelope 
of the reflected rays is called the caustic of the curve with respect to the point. 
Show that the caustic of a circle with respect to a point on its circumference is a 
cardioid. 

16. The curve which is the envelope of the characteristic lines, that is, of the 
rulings, on the developable surface (43) is called the cuspidal edge of the surface. 
Show that the equations of this curve may be found parametrically in terms of the 
parameter of (43) by solving simultaneously 

Ax + By + Cz + D = 0, A'x + B'y + C'z + If = 0, A"x + B"y + C"z + D" = 

for x, y, z. Consider the exceptional cases of cones and cylinders. 

17. The term " developable " signifies that a developable surface may be developed 
or mapped on a plane in such a way that lengths of arcs on the surface become equal 
lengths in the plane, that is, the map may be made without distortion of size or 
shape. In the case of cones or cylinders this map may be made by slitting the cone 
or cylinder along an element and rolling it out upon a plane. What is the analytic 
statement in this case ? In the case of any developable surface with a cuspidal 
edge, the developable surface being the locus of all tangents to the cuspidal edge, 



PAETIAL DIFFEEEXTIATIOX ; IMPLICIT 143 

the length of arc upon the surface may be written as da 2 = (dt + ds) 2 + t 2 ds 2 /B' 2 , 
where s denotes arc measured along the cuspidal edge and t denotes distance along 
the tangent line. This form of da 2 may be obtained geometrically by infinitesimal 
analysis or analytically from the equations 

x=f(s) + tf'(s), y = g(s) + tg'(s), z = h(s) + th'(s) 

of the developable surface of which x =/(s), y = g(s), z = h(s) is the cuspidal edge. 
It is thus seen that da' 2 is the same at corresponding points of all developable sur- 
faces for which the radius of curvature R of the cuspidal edge is the same function 
of s without regard to the torsion ; in particular the torsion may be zero and the 
developable may reduce to a plane. 

18. Let the line x = az + b, y = cz + d depend on one parameter so as to gen- 
erate a 'ruled surface. By identifying this form of the line with (43) obtain by 
substitution the conditions 

Aa + Be + C = 0, A'a + B'c + C = ^ Aa' + Be' = , I of c' I 
Ab + Bd + B = 0, A'b + B'd + B' = ° r Ab' + Bd' =0 ° r W d'\~ 

as the condition that the line generates a developable surface. 

68. More differential geometry. The representation 

F(x,y,z) = 0, or z=f(x,y) (46) 

or x = <£ (u, v), y = xl/(ii, v), z = x( u > v ) 

of a surface may be taken in the unsolved, the solved, or the parametric 
form. The parametric form is equivalent to the solved form provided 
u, v be taken as x, y. 

dz 
1 Cx 

is adopted for the derivatives of z with respect to x and y. The applica- 
tion of Taylor's Formula to the solved form gives 

As = p h + qk + J (rh 2 + 2 shk + tk 2 ) + • • ■ (47) 

with h = Ax, k = Ay. The linear terms ph -f qk constitute the differ- 
ential dz and represent that part of the increment of z which would be 
obtained by replacing the surface by its tangent plane. Apart from 
infinitesimals of the third order, the distance from the tangent plane up 
or down to the surface along a parallel to the 3-axis is given by the 
quadratic terms ^(rh 2 -\- 2 shk -f tk 2 ). 

Hence if the quadratic terms at any point are a positive definite form 
(§ 55), the surface lies above its tangent plane and is concave up; but 
if the form is negative definite, the surface lies below its tangent plane 
and is concave down or convex up. If the form is indefinite but not 
singular, the surface lies partly above and partly below its tangent 
plane and may be called concavo-convex, that is, it is saddle-shaped. If 
the form is singular nothing can be definitely stated. These statements 



The notation 






cz C 2 Z 
cy dx 2 


C 2 Z 

S = r, c 5 

oxvy 


Crz 



lU DIFFERENTIAL CALCULUS 

are merely generalizations of those of § 55 made for the case where the 
tangent plane is parallel to the cc?/-plane. It will be assumed in the 
further work of these articles that at least one of the derivatives r, s, t 
is not 0. 

To examine more closely the behavior of a surface in the vicinity of 
a particular point upon it, let the a^-plane be taken in coincidence with 
the tangent plane at the point and let the point be taken as origin. 
Then Maclaurin's Formula is available. 



z = i (i-x 2 + 2 sxy + ty 2 ) + terms of higher order 
= i p 2 (r cos* 2 + 2 s sin cos 6 + t sin 2 0) + higher terms, ^ ' 

where (p, 6) are polar coordinates in the #2/-plane. Then 

1 d 2 z T /dz\ 2 li 

j = rcos 2 + 2 s sinOcos + tsiii 2 = -^ 2 + [I'+fj-J (49) 

is the curvature of a normal section of the surface. The sum of the 
curvatures in two normal sections which are in perpendicular planes 
may be obtained by giving the values and 6 -\- \ir. This sum 
reduces to r + t and is therefore independent of 6. 

As the sum of the curvatures in two perpendicular normal planes is 
constant, the maximum and minimum values of the curvature will be 
found in perpendicular planes. These values of the curvature are called 
the principal values and their reciprocals are the principal radii of 
curvature and the sections in which they lie are the principal sections. 
If s = 0, the principal sections are = and 6 = \ it ; and conversely 
if the axes of x and y had been chosen in the tangent plane so as to be 
tangent to the principal sections, the derivative s would have vanished. 
The equation of the surface would then have taken the simple form 

z = \ (rx 2 + ty 2 ) + higher terms. (50) 

The principal curvatures would be merely r and t, and the curvature 
in any normal section would have had the form 

1 cos 2 sin 2 (9 . 

- = — h — — = ;• cos- + t sin 2 6. 

A h x A 2 

If the two principal curvatures have opposite signs, that is, if the 
signs of r and t in (50) are opposite, the surface is saddle-shaped. 
There are then two directions for which the curvature of a normal sec- 
tion vanishes, namely the directions of the lines 

6 = ± tan" 1 V— A 2 /A 1 or -s/\r\x = ± V[7j y. 

These are called the asymptotic directions. Along these directions the 
surface departs from its tangent plane by infinitesimals of the third 



PARTIAL DIFFERENTIATION; IMPLICIT 145 

order, or higher order. If a curve is drawn on a surface so that at each 
point of the curve the tangent to the curve is along one of the asymp- 
totic directions, the curve is called an asymptotic curve or line of the 
surface. As the surface departs from its tangent plane by infinitesimals 
of higher order than the second along an asymptotic line, the tangent 
plane to a surface at any point of an asymptotic line must be the oscu- 
lating plane of the asymptotic line. 

The character of a point upon a surface is indicated by the Dupin 
indicatrix of the point. The indicatrix is the conic 

? + £=■*. ci.a = l(rx* + tf), (51) 

which has the principal directions as the directions of its axes and the 
square roots of the absolute values of the principal radii of curvature 
as the magnitudes of its axes. The conic may be regarded as similar to 
the conic in which a plane infinitely near the tangent plane cuts the 
surface when infinitesimals of order higher than the second are neg- 
lected. In case the surface is concavo-convex the indicatrix is a hyper- 
bola and should be considered as either or both of the two conjugate 
hyperbolas that would arise from giving z positive or negative values 
in (ol). The point on the surface is called elliptic, hyperbolic, or 
parabolic according as the indicatrix is an ellipse, a hyperbola, or a pair 
of lines, as happens when one of the principal curvatures vanishes. 
These classes of points correspond to the distinctions definite, indefinite,. 
and singular applied to the quadratic form rh 2 + 2 shk -f- tit?. 

Two further results are noteworthy. Any curve drawn on the surface 
differs from the section of its osculating plane with the surface by 
infinitesimals of higher order than the second. For as the osculating 
plane passes through three consecutive points of the curve, its inter- 
section with the surface passes through the same three consecutive 
points and the two curves have contact of the second order. It follows 
that the radius of curvature of any curve on the surface is identical 
with that of the curve in which its osculating plane cuts the surface. 
The other result is Meusnier's Theorem : The radius of curvature of an 
oblique section of the surface at any point is the projection upon the 
plane of that section of the radius of curvature of the normal section 
which passes through the same tangent line. In other words, if the 
radius of curvature of a normal section is known, that of the oblique 
sections through the same tangent line may be obtained by multiplying 
it by the cosine of the angle between the plane normal to the surface 
and the plane of the oblique section. 



146 DIFFERENTIAL CALCULUS 

The proof of Meusnier's Theorem may be given by reference to (48). Let the 
x-axis in the tangent plane be taken along the intersection with the oblique plane. 
Neglect infinitesimals of higher order than the second. Then 

y = (x) = | ax 2 , z-\ (rx 2 + 2 sxy + ty 2 ) = \ rx 2 (48') 

will be the equations of the curve. The plane of the section is az — ry = 0, as may 
be seen by inspection. The radius of curvature of the curve in this plane may be 
found at once. For if u denote distance in the plane and perpendicular to the 
x-axis and if v be the angle between the normal plane and the oblique plane 

az — ry = 0, 

u = z sec v = y esc v — \ r sec v • x 2 = I a esc v • x 2 . 

The form u — \r sec v ■ x 2 gives the curvature as r sec v. But the curvature in the 
normal section is r by (48'). As the curvature in the oblique section is sec v times 
that in the normal section, the radius of curvature in the oblique section is cos v 
times that of the normal section. Meusnier's Theorem is thus proved. 

69. These investigations with a special choice of axes give geometric proper- 
ties of the surface, but do not express those properties in a convenient analytic 
form ; for if a surface z = f(x, y) is given, the transformation to the special axes 
is difficult. The idea of the indicatrix or its similar conic as the section of the 
surface by a plane near the tangent plane and parallel to it will, however, deter- 
mine the general conditions readily. If in the expansion 

Az - dz = \ (rh 2 + 2 shk + tk 2 ) = const. (52) 

the quadratic terms be set equal to a constant, the conic obtained is the projection 
of the indicatrix on the X2/-plane, or if (52) be regarded as a cylinder upon the 
x?/-plane, the indicatrix (or similar conic) is the intersection of the cylinder with 
the tangent plane. As the character of the conic is unchanged by the projection, 
the point on the surface is elliptic if s 2 < rt, hyperbolic if s 2 > rt, and parabolic if 
s 2 = rt. Moreover if the indicatrix is hyperbolic, its asymptotes must project into the 
asymptotes of the conic (52), and hence if dx and dy replace h and fc, the equation 

rdx 2 + 2 sdxdy + tdy 2 = (53) 

may be regarded as the differential equation of the projection of the asymptotic lines 
on the xy-plane. If r, s, t be expressed as functions f^if^fyy of i x -> V) anc ^ ( 53 ) De 
factored, the integration of the two equations M(x, y)dx + N (x, y)dy thus found 
will give the finite equations of the projections of the asymptotic lines and, taken 
with the equation z =f(x, y), will give the curves on the surface. 

To find the lines of curvature is not quite so simple ; for it is necessary to deter- 
mine the directions which are the projections of the axes of the indicatrix, and 
these are not the axes of the projected conic. Any radius of the indicatrix may 
be regarded as the intersection of the tangent plane and a plane perpendicular to 
the x?/-plane through the radius of the projected conic. Hence 

z-z =p{x-x ) + q(y- y ), (x - x ) k = (y - y ) h 

are the two planes which intersect in the radius that projects along the direction 
determined by h, k. The direction cosines 

h : k :ph + qk 



■s/h 2 + k 2 + (ph + qk) 2 



and h :k:0 (54) 



PAETIAL DIFFEEEXTIATIOX ; IMPLICIT 147 

are therefore those of the radius in the indicatrix and of its projection and they 
determine the cosine of the angle <p between the radius and its projection. The 
square of the radius in (52) is 

h 2 + k 2 , and (h 2 + A; 2 ) sec 2 = h 2 + k 2 + (ph + qk) 2 

is therefore the square of the corresponding radius in the indicatrix. To deter- 
mine the axes of the indicatrix, this radius is to be made a maximum or minimum 
subject to (52). With a multiplier X, 

h + ph + qk + \ (rh + sk) = 0, k + ph + qk + X (sh + tk) = 
are the conditions required, and the elimination of X gives 

h 2 [s (1 + p 2 ) - pqr] + hk [t (1 + p 2 ) - r (1 + q 2 )] - k 2 [t (1 + q 2 ) - pqt] = 
as the equation that determines the projection of the axes. Or 
(1 + P 2 ) dx + pqdij _ pgdx + (l + q 2 )dy 
rdx + sdy sdx + tdy 

is the differential equation of the projected lines of curvature. 

In addition to the asymptotic lines and lines of curvature the geodesic or shortest 
lines on the surface are important. These, however, are better left for the methods 
of the calculus of variations (§ 159). The attention may therefore be turned to 
finding the value of the radius of curvature in any normal section of the surface. 

A reference to (48) and (49) shows that the curvature is 

1 _ 2 z _ rh 2 + 2 shk + tk 2 _ rh 2 + 2 shk + tk 2 
jK - ^ - ^ ~ h 2 + k 2 

in the special case. But in th e general ca se the normal distance to the surface is 
(Az — dz) cos 7, with sec y = Vl + p 2 + 5 2 , instead of the 2 z of the special case, and 
the radius p 2 of the special case becomes p 2 sec 2 4> = h 2 + A; 2 + (ph + qk) 2 in the 
tangent plane. Hence 

1 2 (Az — dz) cos 7 rl 2 + 2 slm + t??* 2 

— = = ==z= — 5 (OO) 

R h 2 + k 2 + (ph + qk) 2 Vl + p 2 + q 2 

where the direction cosines I. m of a radius in the tangent plane have been intro- 
duced from (54), is the general expression for the curvature of a normal section. 
The form 

1 rh 2 + 2 shk + tk 2 1 /Ka/X 

— = — , (oo ) 

B h 2 + k 2 + (ph + qk) 2 Vl + p 2 + q 2 

where the direction h, k of the projected radius remains, is frequently more con- 
venient than (56) which contains the direction cosines I, m of the original direction 
in the tangent plane. Meusnier's Theorem may now be written in the form 

cos v rl 2 + 2 slm + tm 2 _ 

= ==- > (o / ) 

R Vl + p 2 + q' 2 

where v is the angle between an oblique section and the tangent plane and where 
I, m are the direction cosines of the intersection of the planes. 

The work here given has depended for its relative simplicity of statement upon 
the assumption of the surface (46) in solved form. It is merely a problem in 
implicit partial differentiation to pass from p, g, r, s, t to their equivalents in terms 
of F' x , Fy, F' z or the derivatives of 0, \j/, x D 7 a i £• 



148 DIFFEKENTIAL CALCULUS 

EXERCISES 

1 y* ■ I ■ / v ^— f 

1. In (49) show — = 1 cos 2 6 + s sin 2 6 and find the directions of 

v ' B 2 2 

maximum and minimum B. If B x and B 2 are the maximum and minimum values 
of R, show 

1 = r + t and = rt— s 2 . 

.Rj .R 2 i^j JB 2 

Half of the sum of the curvatures is called the mean curvature ; the product of the 
curvatures is called the total curvature. 

2. Find the mean curvature, the total curvature, and therefrom (by construct- 
ing and solving a quadratic equation) the principal radii of curvature at the origin : 

(a) z = xy, (/3) z = x 2 + xy + y 2 , (7) z = x(x + y). 

3. In the surfaces (a) z = xy and (/3) z = 2x 2 + y 2 find at (0, 0) the radius of 
curvature in the sections made by the planes 

{a) x + y = 0, (j8) x + y + z = 0, (7) £ + ?/ + 2 2 = 0, 

(d) x-2y = 0, {e) x-2y + z = 0, (r) x + 2y + |z = 0. 

The oblique sections are to be treated by applying Meusnier's Theorem. 

4. Find the asymptotic directions at (0, 0) in Exs. 2 and 3. 

5 . Show that a developable surface is everywhere parabolic, that is, that rt — s 2 = 
at every point ; and conversely. To do this consider the surface as the envelope of 
its tangent plane z - p x - q y = z - p x - q y , where p , q , x , y , z are func- 
tions of a single parameter a. Hence show 

j(P0 1 _%\ =0 = {rt _ s 2 ) and j( P0i Z - JW) - Q^o \ = y (S 2 _ H) 

The first result proves the statement ; the second, its converse. 

6. Find the differential equations of the asymptotic lines and lines of curvature 
on these surfaces : 

(a) z = xy, (|8) z = tan-^y/x), (7) z 2 + y 2 = coshx, (5) xyz = 1. 

7. Show that the mean curvature and total curvature are 

■) 2 ) — 2pqs 1 rt — s 2 



1 /I _1\ = r(l + g 2 ) + t(l+p s 

2 \E 1 E 2 / 2 (1 + p 2 + g 5 



2\B X E 2 / 2(l+p 2 + g 2 )t «i^ 2 (l+p 2 +g 2 ) 2 

8. Find the principal radii of curvature at (1, 1) in Ex. 6. 

9. An umbilic is a point of a surface at which the principal radii of curvature 
(and hence all radii of curvature for normal sections) are equal. Show that the 

conditions are = — = for an umbilic, and determine the umbilics of 

1 + p 2 pq 1 + q 2 

the ellipsoid with semiaxes a, b, c. 



CHAPTER VI 

COMPLEX NUMBERS AND VECTORS 

70. Operators and operations. If an entity u is changed into an 
entity v by some law, the change may be regarded as an operation per- 
formed upon u, the operand, to convert it into v ; and if / be introduced 
as the symbol of the operation, the result may be written as v —fu. 
For brevity the symbol f is often called an operator. Various sorts 
of operand, operator, and result are familiar. Thus if wis a positive 
number ?i, the application of the operator V gives the square root ; if u 
represents a range of values of a variable x, the expression f(x) or fx 
denotes a function of x ; if u be a function of x, the operation of dif- 
ferentiation may be symbolized by D and the result Da is the deriva- 
tive ; the symbol of definite integration I (#) d* converts a function 

u (x) into a number ; and so on in great variety. 

The reason for making a short study of operators is that a consider- 
able number of the concepts and rules of arithmetic and algebra may 
be so defined for operators themselves as to lead to a calculus of opera- 
tions which is of frequent use in mathematics ; the single application to 
the integration of certain differential equations (§ 95) is in itself highly 
valuable. The fundamental concept is that of a product : If u is oper- 
ated upon by f to givefu = v and if v is operated upon by g to give gv = w, 

so that 

fu = v, gv = gfu = iv, gfu = u; (1) 

then the operation indicated as gf which converts u directly into w is 
called the product of f by g. If the functional symbols sin and log be 
regarded as operators, the symbol log sin could be regarded as the 
product. The transformations of turning the £?/-plane over on the 
^-axis, so that x' = x, y' = — y, and over the ?/-axis, so that x' = — x, 
y' = y, may be regarded as operations ; the combination of these opera- 
tions gives the transformation x' = — x, y' = — y, which is equivalent 
to rotating the plane through 180° about the origin. 

The products of arithmetic and algebra satisfy the commutative law 
gf = fg, that is, the products of g by / and of / by g are equal. This 
is not true of operators in general, as may be seen from the fact that 

149 



150 DIFFERENTIAL CALCULUS 

log sin x and sin log x are different. Whenever the order of the factors 
is immaterial, as in the case of the transformations jnst considered, the 
operators are said to be commutative. Another law of arithmetic and 
algebra is that when there are three or more factors in a product, the 
factors may be grouped at pleasure without altering the result, that is, 

H'jf) = (M)f=hgf- (2) 

This is known as the associative law and operators which obey it are 
called associative. Only associative operators are considered in the 
work here given. 

For the repetition of an operator several times 

ff = A fff = f, f m f n =f m+n , (3) 

the usual notation of powers is used. The law of indices clearly holds ; 
for f m + n means that / is applied m -\- n times successively, whereas 
f m f n means that it is applied n times and then m times more. Xot 
applying the operator / at all would naturally be denoted by /°, so that 
f°u = u and the operator f° would be equivalent to multiplication by 1 ; 
the notation f° = 1 is adopted. 

If for a given operation / there can be found an operation g such 
that the product fg =f° = l is equivalent to no operation, then g is 
called the inverse of / and notations such as 

fg = l, ? =/-i = i, //-i =/ I = l (4) 

are regularly borrowed from arithmetic and algebra. Thus the inverse 
of the square is the square root, the inverse of sin is sin -1 , the inverse 

of the logarithm is the exponential, the inverse of D is / . Some oper- 
ations have no inverse ; multiplication by is a case, and so is the 
square when applied to a negative number if only real numbers are 
considered. Other operations have more than one inverse ; integra- 
tion, the inverse of D, involves an arbitrary additive constant, and the 
inverse sine is a multiple valued function. It is therefore not always 
true that f~ 1 f= 1, but it is customary to mean by f~ 1 that particular 
inverse of / for which f- 1 f = ff- 1 = l. Higher negative powers are 
defined by the equation f~ n = (/ -1 ) n , and it readily follows that 
f n f~ n = l, as may be seen by the example 

ff~ % =M/-f- 1 )f-V- 1 =f(f-f- 1 )/- 1 =ff- 1 = l. 

The law of indices f m f n =f m + n also holds for negative indices, except 
in so far as f~\f may not be equal to 1 and may be required in the 
reduction of f m f n to f m+n . 



COMPLEX NUMBERS AXD VECTORS 151 

If u, v, and u -f- v are operands for the operator /and if 

f(u + v)=fu+fv, (5) 

so that the operator applied to the sum gives the same result as the 
sum of the results of operating on each operand, then the operator 
/ is called linear or distributive. If / denotes a function such that 
f(x 4- y) =f(x) +/(//), it has been seen (Ex. 9, p. 45) that / must be 
equivalent to multiplication by a constant and fx = Cx. For a less 
specialized interpretation this is not so ; for 



D(u+ v) 



Du + Dv and j (u + v) = I u + I i 



are two of the fundamental formulas of calculus and show operators 
which are distributive and not equivalent to multiplication by a constant. 
Nevertheless it does follow by the same reasoning as used before (Ex. 9, 
p. 45), th&tfnu = nfu if /is distributive and if n is a rational number. 
Some operators have also the property of addition. Suppose that u 
is an operand and/, g are operators such that/it and gu are things that 
may be added together as fu + gu, then the sum of the operators, /+ fj, 
is defined by the equation (/ -f g)u =fu -{- gu. If furthermore the 
operators /, g, h are distributive, then 

h(f+g) = hf+hg and (f + g) h=fh + gh, (6) 

and the multiplication of the operators becomes itself distributive.- To 
prove this fact, it is merely necessary to consider that 

4 [(/+ 9) "] = *(A + H = *> + ^ 
and (/ + g) (hu) = fh u + gh u. 

Operators which are associative, commutative, distributive, and which 
admit addition may be treated algebraically, in so far as polynomials are 
concerned, by the ordinary algorisms of algebra ; for it is by means 
of the associative, commutative, and distributive laws, and the law of 
indices that ordinary algebraic polynomials are rearranged, multiplied 
out, and factored. Xow the operations of multiplication by constants 
and of differentiation or partial differentiation as applied' to a function 
of one or more variables x, y, z, ■ • • do satisfy these laws. For instance 

c(Du) =D(cu), DJ) y u = D v D x u, (D x + D y )D z u = D x D z u + D y D z u. (7) 
Hence, for example, if y be a function of x, the expression 

D n y + a x D«-Hj + • • • + <i n _J>y + ajj, 
where the coefficients a are constants, may be written as 

(pn + apn -i + ... + Cbn _ iD + ,g y (8) 



152 DIFFERENTIAL CALCULUS 

and may then be factored into the form 

[(0 - o0(2> - «Q ...(D- a._J(p - a,)-]y, (8') 

where a v a 2 , • • -, a n are the roots of the algebraic polynomial 
x n + ajfi-i _| ^_ C(n _ iX + an = 0. 

EXERCISES 

1. Show that (fgh)- 1 = /i -1 ^ -1 / -1 , that is, that the reciprocal of a product of 
operations is the product of the reciprocals in inverse order. 

2. By definition the operator gfg- 1 is called the transform of /by g. Show 
that (a) the transform of a product is the product of the transforms of the factors 
taken in the same order, and (/3) the transform of the inverse is the inverse of the 
transform. 

3. If s y£ 1 but s 2 = 1, the operator s is by definition said to be involutory. Show 
that (a) an involutory operator is equal to its own inverse ; and conversely (/3) if 
an operator and its inverse are equal, the operator is involutory ; and (7) if the 
product of two involutory operators is commutative, the product is itself involu- 
tory ; and conversely (5) if the product of two involutory operators is involutory, 
the operators are commutative. 

4. If /and g are both distributive, so are the products fg and gf. 

5. If /is distributive and n rational, show fnu = nfu. 

6. Expand the following operators first by ordinary formal multiplication and 
second by applying the operators successively as indicated, and show the results 
are identical by translating both into familiar forms. 

(a) (D-l)(D-2)y, Ans. pL-3^ + 2y, 

dx 2 dx 

(P) (B-l)J)(B + l)y, (7) D(D-2)(D+l)(D + 3)y. 

7. Show that (D— a)\ e ax \ e~ ax Xdx = X, where X is a function of x, and 
hence infer that e ax I e- ax (*)dx is the inverse of the operator (D — a) (*). 

8. Show that D (e ax y) = e ax (D + a) y and hence generalize to show that if 
P (D) denote any polynomial in D with constant coefficients, then 

P (D) • e ax y = e ax P (D + a) y. 
Apply this to the following and check the results. 

(a) (Z>2 _ 3D + 2)e»*y = e 2x (D- 2 + D)y = e 2x (^ + ^)' 
(|8) (D 2 - 3 D - 2) e x y, (7) (D 3 - 3 D + 2) e*y. 

9. If y is a function of x and x = e f show that 

I) x y = e-Wty, D' 2 y = er**B t (D t - l)y, • • •, D£y = e-P*D t {D t - 1) • • • (A-JP + l)tf. 

10. Is the expression (hD x + feD y ) n , which occurs in Taylor's Formula (§ 54), 
the nth power of the operator hD x + kD v or is it merely a conventional symbol ? 
The same question relative to (xD x + yD y ) k occurring in Euler's Formula (§ 53) ? 



COMPLEX NUMBERS AND VECTORS 153 

71. Complex numbers. In the formal solution of the equation 
ax 2 + bx + c = 0, where b 2 < 4 ac, numbers of the form m + n V— 1, 
where m and n are real, arise. Such numbers are called complex or 
imaginary ; the part m is called the real part and ?iV- 1 the jp?«re 
imaginary part of the number. It is customary to write V— 1 = 1 and 
to treat i as a literal quantity subject to the relation i 2 = —1. The defini- 
tions for the equality, addition, and multiplication of complex num- 

a -\-bi = c -\- di if and only if a = c, b = (7, 

[a + bi] + [> + df] = (a + c) + (b + d) *, (9) 

[a + W] [c + di] = (ac — bd) + (ad -j- 5c) i. 

It readily follows that the commutative, associative, and distributive 
lairs hold in the domain of complex numbers, namely, 

a-+p = p + a, (a + 0) + y = a + (0 + y), 

«£ = /8a, (a£) y - a (J3y), (10) 

« (P + r) = «£ + *y> (* + /*) y = «y + £y> 

where Greek letters have been used to denote complex numbers. 
Division is accomplished by the method of rationalization. 

a -f- bi _ a -\- bi c — di _ (ac -f- &d) + (be — ad) i 

c + di ~ e + di c — di ~ r + ^ 2 ^ 

This is always possible except when c 2 + <r/' 2 = 0, that is, when both c 
and d are 0. A complex number is defined as when and only when 
its real and pure imaginary parts are both zero. With this definition 
has the ordinary properties that a -f = a and a = and that a/0 is 
impossible. Furthermore if a product aft vanishes, either a or ft vanishes. 
For suppose 

[a + bi] [c + di] = (ac — bd) + (ad + be) i = 0. 
Then ac - bd = and ad + be = 0, (12) 

from which it follows that either a = b = or c = d = 0. From the 
fact that a product cannot vanish unless one of its factors vanishes 
follow the ordinary laws of cancellation. In brief, all the elementary 
laws of real algebra hold also for the algebra of complex numbers. 

By assuming a set of Cartesian coordinates in the £c?/-plane and asso- 
ciating the number a -\- bi to the point, (a, b), a graphical representation 
is obtained which is the counterpart of the number scale for real num- 
bers. The point (a, b) alone or the directed line from the origin to the 
point (a, b) may be considered as representing the number a + bi. 
If OP and OQ are two directed lines representing the two numbers 
a -{-bi and c + di, a reference to the figure shows that the line which 




154 DIFFERENTIAL CALCULUS 

represents the sum of the numbers is OR, the diagonal of the parallelo- 
gram of which OP and OQ are sides. Thus the geometric law for adding 
complex numbers is the same as the law for compounding forces and is 
knoivn as the parallelogram law. A segment i5 of a line possesses 
magnitude, the length A B, and direction, the 

direction of the line AB from A to B. A v n ^ t (a+c,b+d) 

quantity which has magnitude and direction is 
called a vector • and the parallelogram law is 
called the lata of vector addition. Complex num- 
bers may therefore be regarded as vectors. 

From the figure it also appears that OQ and PR have the same mag- 
nitude and direction, so that as vectors they are equal although they 
start from different points. As OP -\- PR will be regarded as equal to 
OP + OQ, the definition of addition may be given as the triangle law 
instead of as the parallelogram law ; namely, from the terminal end P 
of the first vector lay off the second vector PR and close the triangle 
by joining the initial end of the first vector to the terminal end R of 
the second. The absolute value of a complex number a -p- hi is the 
magnitude of its vector OP and is equal to Va 2 -f- b 2 , the square root of 
the sum of the squares of its real part and of the coefficient of its pure 
imaginary part. The absolute value is denoted by \a -f bi\ as in the case 
of reals. If a and (3 are two complex numbers, the rule |tf|-f-|/3| = |# + /3| 
is a consequence of the fact that one side of a triangle is less than the 
sum of the other two. If the absolute value is given and the initial end 
of the vector is fixed, the terminal end is thereby constrained to lie 
upon a circle concentric with the initial end. 

72. When the complex numbers are laid off from the origin, polar 
coordinates may be used in place of Cartesian. Then 



r = Vcr + lr, <£ = tan -1 &/«*, a = r cos <£, b = r sin <f> 

\ ) 
and a + ib = r(cos cj> + i sin <f>). 

The absolute value r is often called the modulus or magnitude of the 
complex number; the angle <f> is called the angle or argument of the 
number and suffers a certain indetermination in that 2 nnr, where n is 
a positive or negative integer, may be added to <f> without affecting the 
number. This polar representation is particularly useful in discussing 
products and quotients. For if 

a = r x (cos ^ + i sin ^), fi = r 2 (cos <£ 2 + i sin <£ 2 ), 

then a(3 = ^[cos (^ + <£ 2 ) + i sin (^ + <£ 2 )], 

* As both cos </> and sin </> are known, the quadrant of this angle is determined. 



COMPLEX NUMBERS AND VECTORS 155 

as may be seen by multiplication according to the rule. Hence the 
magnitude of a product is the product of the magnitudes of the factoids, 
and the angle of a product is the sum of the angles of the factors; the 
general rule being proved by induction. 

The interpretation of multiplication by a complex number as an oper- 
ation is illuminating. Let ft be the multiplicand and a the multiplier. 
As the product a/3 has a magnitude equal to the product of the magni- 
tudes and an angle equal to the sum of the angles, the factor a used as 
a multiplier may be interpreted as effecting the rotation of (3 through 
the angle of a and the stretching of (3 in the ratio \a\:l. From the 
geometric viewpoint, therefore, multiplication by a complex number is 
an operation of rotation and stretching in the plane. In the case of 
a = cos <{> + i sin <£ with r = 1, the operation is only of rotation and 
hence the factor cos <j> -\- i sin <f> is often called a cyclic factor or versor. 
In particular the number i = V— 1 will effect a rotation through 90° 
when used as a multiplier and is known as a quadrantal versor. The 
series of powers i, i 2 = — 1, i 3 = — i, i 4 = 1 give rotations through 90°, 
180°, 270°, 360°. This fact is often given as the reason for laying off 
pure imaginary numbers bi along an axis at right angles to the axis 
of reals. 

As a particular product, the nth power of a complex number is 

a n = (a + ib) n = [r (cos <f> + i sin <£)]" = r n (cos n<j> -f- i sin ncf>) ; (15) 

and (cos cf> -f i sin <£)' 1 = cos ?i<f> -f- i sin ?icf>, (15') 

which is a special case, is known as De Moivre's Theorem and is of use 
in evaluating the functions of n<f> ; for the binomial theorem may be 
applied and the real and imaginary parts of the expansion may be 
equated to cos n$ and sin n§. Hence 

n (n — 1) „ . ' 

cos n<f> = eos n cf> ^-— — - cos n ~ 2 cf> snr<£ 

,»(n-l)(ft-2)(»-3) ... 
+ — ^-j^ — ~ — cos" - 4 <£ sm 4 <£ (16) 

n(n — l)(n — 2) 
sin n<j> = n cos n ~ 1 <f> sin <£ ; ^j } - cos n-3 <£ sin 3 <£ -\ . 

As the nth. root -\/a of a must be a number which when raised to the 
nth power gives a, the nth root may be written as 

\a = -\/r(cos <f>/n -f i sin 4>/n). (17) 

The angle <f>, however, may have an} T of the set of values 

<fc 4> + 2tt, ^ + 4tt, •-., * + 2(n-l)w, 



156 DIFFEEENTIAL CALCULUS 

and the nth. parts of these give the n different angles 

<£ <£ 2tt <j> 4tt <fr 2(tt-l)7r , 

n n n n n n n 

Hence there may be found just n different nth roots of any given com- 
plex number (including, of course, the reals). 

The roots of unity deserve mention. The equation x n = 1 has in the real domain 
one or two roots according as n is odd or even. But if 1 be regarded as a complex 
number of which the pure imaginary part is zero, it may be represented by a point 
at a unit distance from the origin upon the axis of reals ; the magnitude of 1 is 1 
and the angle of 1 is 0, 2 7r, • • • , 2 (n — 1) it. The nth roots of 1 will therefore have 
the magnitude 1 and one of the angles 0, 2 ir/n, ■ • • , 2 (n — 1) ir/n. The n nth roots 
are therefore 

2tt , . . 2tt . 4tt , . . 4tt 

1, a = cos Msin — , a 1 = cos Msin — > '"> 

n n n n 

2(n-l)ir . . 2(n-l)7r 

a" - ! = COS — 1- l SID — '— , 

n n 

and may be evaluated with a table of natural functions. Now x n — 1 =0 is factor- 
able as (x — l)(x n_1 + x n ~ 2 + • • • + x + 1) = 0, and it therefore follows that the 
nth roots other than 1 must all satisfy the equation formed by setting the second 
factor equal to 0. As a in particular satisfies this equation and the other roots are 
a 2 , • • • , a M_1 , it follows that the sum of the n nth roots of unity is zero. 



EXERCISES 

1. Prove the distributive law of multiplication for complex numbers. 

2. By definition the pair of imaginaries a + bi and a— bi are called conjugate 
imaginaries. Prove that (a) the sum and the product of two conjugate imaginaries 
are real ; and conversely (/3) if the sum and the product of two imaginaries are both 
real, the imaginaries are conjugate. 

3. Show that if P(x, y) is a symmetric polynomial in x and y with real coeffi- 
cients so that P(x, y) = P(y', x), then if conjugate imaginaries be substituted forx 
and y, the value of the polynomial will be real. 

4. Show that if a + bi is a root of an algebraic equation P(x) = with real 
coefficients, then a — bi is also a root of the equation. 

5. Carry out the indicated operations algebraically and make a graphical repre- 
sentation for every number concerned and for the answer : 

(a) (1 + i) 3 , ' (j8) (1 + V3 (1 - i), ( 7 ) (3 + V^2) (4 + V^~5), 

(J) i±i, ( e) l±M, (n 



1 - * 1-iVs V2 - i V3 

(1-z) 2 /m 1 , 1 / x / -l + V_3 \3 



^JTTif' W <T+^ + (T^' <•> 



Plot and find the modulus and angle in the following cases: 

(a) -2, ( j8 )_2V=T, (7)3 + 4*, (5) \ - \ V^3. 



COMPLEX NUMBERS AND VECTORS 157 

7. Show that the modulus of a quotient of two numbers is the quotient of the moduli 
and that the angle is the angle of the numerator less that of the denominator. 

8. Carry out the indicated operations trigonometrically and plot: 

(a) The examples of Ex. 5, (/3) VT+i Vl — t, (7) V- 2 + 2V3i, 

(5) ( vm + vr^Q 2 , (e) V V2 + v^2, (n y2 + 2V3i, 

( v ) </lQ (cos 200° + i sin 200°), (0) \^- 1, (1) \^8T. 

9. Find the equations of analytic geometry which represent the transforma- 
tion equivalent to multiplication by a = — 1 + V— 3. 

10. Show that \z — a\ = r, where z is a variable and a a fixed complex number, 
is the equation of the circle (x — a) 2 + (y — b) 2 = r 2 . 

1 1 . Find cos 5 x and cos 8 x in terms of cos x, and sin 6 x and sin 7 x in terms of 
sin j. 

12. Obtain to four decimal places the five roots Vl. 

13. If z = x + iy and z* — x' + iy\ show that z' — (cos cp — i sin <p) z — a is the 
formula for shifting the axes through the vector distance a — a + ib to the new 
origin (a, b) and turning them through the angle 0. Deduce the ordinary equa- 
tions of transformation. 

14. Show that \z — a\ = k\z — /3], where k is real, is the equation of a circle ; 
specify the position of the circle carefully. Use the theorem : The locus of points 
whose distances to two fixed points are in a constant ratio is a circle the diameter 
of which is divided internally and externally in the same ratio by the fixed points. 

15. The transformation z' — , where a, 6, c, d are complex and ad—bc^. 0, 

cz + d 
is called the general linear transformation of z into z'. Show that 

ea' + d 
cp + d 

Hence infer that the transformation carries circles into circles, and points which 
divide a diameter internally and externally in the same ratio into points which 
divide some diameter of the new circle similarly, but generally with a different ratio. 

73. Functions of a complex variable. Let z = x 4- iy be a complex 
variable representable geometrically as a variable point in the sc^-plane, 
which may be called the complex plane. As z determines the two real 
numbers x and y, any function F(x, y) which is the sum of two single 
valued real functions in the form 

F{x, y) = X(x. y) + iY(x, y) = R (cos $ + i sin $) (19) 

will be completely determined in value if z is given. Such a function 
is called a complex function (and not a function of the complex variable, 
for reasons that will appear later). The magnitude and angle of the 
function are determined by 

X . Y 



\z' — a'\ = k\z' — /S'l becomes \z— a\ 



P\- 



R = VX 2 + F 2 , cos $ = - , sin $ = - . (20) 

R R 



158 DIFFERENTIAL CALCULUS 

The function F is continuous by definition when and only when both 
X and Y are continuous functions of (x, y) ; R, is then continuous in 
(x, y) and F can vanish only when R = ; the angle <I> regarded as a 
function of (x, y) is also continuous and determinate (except for the 
additive 2 nir) unless R = 0, in which case X and Y also vanish and the 
expression for $ involves an indeterminate form in two variables and 
is generally neither determinate nor continuous (§ 44). 

If the derivative of F with respect to z were sought for the value 
z = a + ib, the procedure would be entirely analogous to that in the 
case of a real function of a real variable. The increment Az = Ax + iAy 
would be assumed for z and AF would be computed and the quotient 
AF/Az would be formed. Thus by the Theorem of the Mean (§ 46), 

AF _ AX + iA Y = (X' r + i r T ) Ax + (X' tt + i Y' y ) A?/ . (2± . 

Az Ax + iAy Ax + iAy ±> \ ) 

where the derivatives are formed for (a, b) and where £ is an infinitesi- 
mal complex number. When Az approaches 0, both Ax and Ay must 
approach without any implied relation between them. In general the 
limit of AF/Az is a double limit (§ 44) and may therefore depend on 
the way in which A^ and Ay approach their limit 0. 

Now if first Ay = and then subsequently Ax = 0, the value of the 
limit of AF/Az is X% + %Y' X taken at the point (a, b) ; whereas if first 
Ax = and then Ay = 0, the value is — %X' y + Y' y . Hence if the limit 
of AF/Az is to be independent of the way in which Az approaches 0, it 
is surely necessary that 

d X .dY .dX BY 

ex ex cy 0y 

dx dY dx dY ' 

or -— = — - and —- = — -—• (22) 

dx cy cy ex v y 

And conversely if these relations are satisfied, then 

AF (dX , .dY\ „ /dY .dX\ , ^ 
Az \dx dx) ^ \dy dy ) b ' 

and the limit is X' x + %Y' X = Y' y — iX y taken at the point (a, b), and is 
independent of the way in which Az approaches zero. The desirability 
of having at least the ordinary functions differentiable suggests the 
definition: A complex function F(x, y)—X(x, y) + iY(x, y) is con- 
sidered as a function of the complex variable z = x + iy when and only 
when X and Y are in general differentiable and satisfy the relations (22). 
In this case the derivative is 



COMPLEX XUMBEES AKD YECTOES 159 

dF cX m dY cY .cX , rtOX 

F'(z) = — = — +i — = - i — ■ (23) 

y clz ex ex cy ex J 

These conditions may also be expressed in polar coordinates (Ex. 2). 

A few words about the function <f>(x, y). This is a multiple valued function of 
the variables (x, y), and the difference between two neighboring branches is the con- 
stant 2 ir. The application of the discussion of § 45 to this case shows at once that, 
in any simply connected region of the complex plane which contains no point (a, b) 
such that R (a, b) = 0, the different branches of $ (x, y) may be entirely separated 
so that the value of <£ must return to its initial value when any closed curve is de- 
scribed by the point (x, y). If, however, the region is multiply connected or contains 
points for which R = (which makes the region multiply connected because these 
points must be cut out), it may happen that there will be circuits for which <£, 
although changing continuously, will not return to its initial value. Indeed if it can 
be shown that $ does not return to its initial value when changing continuously as 
(x, y) describes the boundary of a region simply connected except for the excised 
points, it may be inferred that there must be points in the region for which R = 0. 

An application of these results may be made to give a very simple demonstration 
of the fundamental theorem of algebra that every equation of the nth degree has at least 
one root. Consider the function 

F(z) = z n + djiP- 1 + • • • + dn-iz + a n = X{x, y) + iY(x, y), 

where X and Y are found by writing z as x + iy and expanding and rearranging. 
The functions X and Y will be polynomials in (x, y) and will therefore be every- 
where finite and continuous in (x, y). Consider the angle $ of F. Then 

$ = ang. of F = ang. of 2» (l + — + ••• + ^— - + — ) = ang. of z n + ang. of (1 + • • •). 

\ z z 71 - 1 Z n J 

Next draw about the origin a circle of radius r so large that 

= K1 { ... [ K-il { >»1 <C> 

f yn —1 ytl 

Then for all points z upon the circumference the angle of F is 

$ = ang. of jP = n (ang. of z) + ang. of (1 + 77), 1 77 1 < e. 

Now let the point (x, y) describe the circumference. The angle of z will change by 
2 7r for the complete circuit. Hence <£> must change by 2 mr and does not return to 
its initial value. Hence there is within the circle at least one point (a, b) for which 
R (a, b) = and consequently for which X(a, b) = and Y(a, b) = and F(a, b) = 0. 
Thus if a = a + #>, then F(a) = and the equation F(z) = is seen to have at 
least the one root a. It follows that z — a is a factor of F (z) ; and hence by induc- 
tion it may be seen that F(z) = has just n roots. 

74. The discussion of the algebra of complex numbers showed how 
the sum, difference, product, quotient, real powers, and real roots of 
such numbers could be found, and hence made it possible to compute 
the value of any given algebraic expression or function of z for a given 
value of z. It remains to show that any algebraic expression in z is 



a 1 


+ • 


• + 


«n-l| 


On 


z 






z n-l\ 


z n 



160 DIFFERENTIAL CALCULUS 

really a function of z in the sense that it has a derivative with respect 
to z, and to find the derivative. Now the differentiation of an algebraic 
function of the variable x was made to depend upon the formulas of dif- 
ferentiation, (6) and (7) of § 2. A glance at the methods of derivation 
of these formulas shows that they were proved by ordinary algebraic 
manipulations such as have been seen to be equally possible with imagi- 
naries as with reals. It therefore may be concluded that an algebraic 
expression in z lias a derivative with respect to z and that derivative 
may be found just as if z were a real variable. 

The case of the elementary functions e z , log z, sin z, cos z, • • • other 
than algebraic is different ; for these functions have not been defined 
for complex variables. Now in seeking to define these functions when z 
is complex, an effort should be made to define in such a way that : 1° 
when z is real, the new and the old definitions become identical ; and 
2° the rules of operation with the function shall be as nearly as possi- 
ble the same for the complex domain as for the real. Thus it would be 
desirable that De z = e z and e z + w = e z e w , when z and w are complex. 
With these ideas in mind one may proceed to define the elementary 
functions for complex arguments. Let 

e z = R (x, y) [cos 3> (x, y) + i sin <E> (x, y)~\. (24) 

The derivative of this function is, by the first rule of (23), 

Be z = — (R cos 3>) + i — (R sin <£) 

ex v J dx K J 

= (R x cos 3> — R sin 3> • <£^) + i (R x sin <£ -J- R cos 3> • <SQ, 

and if this is to be identical with e z above, the equations 

R ' cos $ — R& r sin <£ = R cos <J> R' = R 

R x sm <3? + R$ x cos 3> = R sm 3> $ x = 

must hold, where the second pair is obtained by solving the first. If 
the second form of the derivative in (23) had been used, the results 
would have been R' y = 0, ^ = 1. It therefore appears that if the 
derivative of e z , however computed, is to be e z , then 

are four conditions imposed upon R and <£. These conditions will be 
satisfied if R = e x and 3> = y.* Hence define 

e z _ e x + iy _ e x ( cos y _j_ i s j n y y (25) 

* The use of the more general solutions R = Ge x , $ = y+C would lead to expressions 
which would not reduce to e x when y = and z = xor would not satisfy e z + w = e z e w . 



■am 



COMPLEX NUMBEES AND VECTOES 161 

With this definition De z is surely e 3 , and it is readily shown that the 
exponential law e z + w = e z e w holds. 

For the special values i iri, tti, 2 iri of z the value of e z is 

e^* = i, e"* = — 1, e 27 "' = l. 

Hence it appears that if 2 niri be added to z, e z is unchanged ; 

e z + 2n ™ = e z , period 2iri. (26) 

Thus in the complex domain e z has the period 2 7ri, just as cos x and 
sin x have the real period 2 it. This relation is inherent ; for 

e yi = cos y + £ sin ?/, e~ ** = cos y — i sin ?/, 

e yi + e -if* _ e ^ _ g-3* 
and cos ?/ = ? sin ?/ = — (2 < ) 

The trigonometric functions of a real variable y may be expressed in 
terms of the exponentials of yi and — yi. As the exponential has been 
defined for all complex values of z, it is natural to use (27) to define 
the trigonometric functions for complex values as 

e zi + e~ si . e zi — e~ zi /0 _ 

cos z = j sin z = — — (2 r) 

With these definitions the ordinary formulas for cos (z + iv), D sin z, • ■ ■ 
may be obtained and be seen to hold for complex arguments, just as the 
corresponding formulas were derived for the hyperbolic functions (§ 5). 
As in the case of reals, the logarithm log z will be defined for com- 
plex numbers as the inverse of the exponential. Thus 

if e z = ic, then log iv = z -f- 2 niri, (28) 

where the periodicity of the function e z shows that the logarithm is not 
uniquely determined but admits the addition of 2 niri to any one of its 
values, just as tan -1 x admits the addition of mr. If w is written as a 
complex number u + iv with modulus r = V^ 2 -f v' 2 and with the angle 
4>, it follows that 

w =u+ iv = r(cos <f> -h i sin <£) == re* 1 = e losr + (pl ; (29) 

and log w = log r + <f>i = log V u 2 -{- v 2 _|_ i tan -1 (v/u) 

is the expression for the logarithm of w in terms of the modulus and 
angle of w ; the 2 niri may be added if desired. 

To this point the expression of a power a b , where the exponent b is 
imaginary, has had no definition. The definition may now be given in 
terms of exponentials and logarithms. Let 

a b _ g&loga or } g a b _ Jj l Q g a> 



162 DIFFEKENTIAL CALCULUS 

In this way the problem of computing a b is reduced to one already 
solved. From the very definition it is seen that the logarithm of a 
power is the product of the exponent by the logarithm of the base, as 
in the case of reals. To indicate the path that has been followed in 
defining functions, a sort of family tree may be made. 

real numbers, x real angles, x 

I I . 

real powers and real trigonometric functions, 
roots of reals, x n cosx, sinx, tair-^-x, • • • 

I ' 1 i ' 

exponentials, logarithms real powers and roots 

of reals, e* logx of imaginaries, z n 



exponentials of imaginaries, e z 

I L , 



l 

logarithms of imaginaries, log z trigonometric functions 

I of imaginaries 

imaginary powers, z a 

EXERCISES 

1. Show that the following complex functions satisfy the conditions (22) and 
are therefore functions of the complex variable z. Find F'(z): 

{a) x 2 - y 2 + 2 ixy, (/3) x 3 - 3 {xy 2 + x 2 - y 2 ) + i (3 xhj - y* - 6 xy), 

(7) ^—,-i~-7 2 > («) logVxM^ + Uan-^, 
x 2 + y 2 x 2 + y 2 x 

( e ) e x cos y + ie x sin y, (f) sin x smb. y + i cos x cosh y. 

2. Show that in polar coordinates the conditions for the existence of F'(z) are 

dX \cY dY ldX . +1 _„, (dX , .dY\. ^ . . . 

= , — = with F'(z) = [ f- i — )(cos0 — isin^). 

dr r g0 dr r d(j> \ cr dr J 

3 . Use the conditions of Ex. 2 to show from D log z = z- 1 that log 2 = log r + 0i. 

4. From the definitions given above prove the formulas 

(a) sin (x + iy) = sin x cosh ?/ + i cos x sinh y, 

(j8.) cos (x + iy) = cos x cosh ?/ — i sin x sinh y, 

. . . , , . N sin 2 x + i sinh 2 ?/ 

(7) tan (x + iy) = 

v ' v ' cos 2 x + cosh 2 y 

5. Find to three decimals the complex numbers which express the values of : 

(a) ei 7 "', (?) e\ (7) ei + i^ (5) e-i-<, 

(e) siniTri, (f),cos«, (77) sin(i + iV-3), (9) tan (- 1 - i), 

(0 log(-l), (k) logi, (X) log(i + fV-8), ( M ) log(-l-i). 

6. Owing to the fact that log a is multiple valued, a b is multiple valued in such 

a manner that any one value may be multiplied by e 2nnbi . Find one value of each 

of the following and several values of one of them : 

3 . + 1 

(a) 2S 08) i\ (7) -W, (8) v5, (e) (1 + W-3)- . 



COMPLEX XUMBEKS AXD VECTORS 

7. Show that Da z = a z \oga when a and z are complex. 



163 



8. Show that (a b ) c = a bc ; and fill in such other steps as may be suggested by 
the work in the text, which for the most part has merely been sketched in a broad 
way. 

9. Show that if f(z) and g(z) are two functions of a complex variable, then 
f(z)±g(z), af(z) with a a complex constant, f(z)g(z), f(z)/g(z) are also func- 
tions of z. 

10. Obtain logarithmic expressions for the inverse trigonometric functions. 
Find sin- 1 /. 

75. Vector sums and products. As stated in § 71, a vector is a quan- 
tity which lias magnitude and direction. If the magnitudes of two 
vectors are equal and the directions of the two vectors are the same, 
the vectors are said to be equal irrespective of the 
position which they occupy in space. The vector 
— a is by definition a vector which has the same 
magnitude as a but the opposite direction. The 
vector ma is a vector which has the same direction 
as a (or the opposite) and is m (or — m) times as 
long. The law of vector or geometric addition is 
the parallelogram or triangle law (§ 71) and is still 
applicable when the vectors do not lie in a plane 
but have any directions in space ; for any two vec- 
tors brought end to end determine a plane in which the construction 
may be carried out. Vectors will be designated by Greek small letters 
or by letters in heavy type. The relations of equality or similarity 
between triangles establish the rules 

a + (3 = f3 + «, « + (0 + y) = (* + P) + y, m (a + 0) = ma + mfi (30) 

as true for vectors as well as for numbers whether real or complex. A 
vector is said to be zero when its magnitude is zero, and it is writ- 
ten 0. From the definition of addition it follows that 
a + = a. In fact as far as addition, subtraction, and 
multiplication by numbers are concerned, vectors obey 
the same formal laics as numbers. 

A vector p ma}' be resolved into components par- 
allel to any three given vectors a, (3, y which are not 
parallel to any one plane. For let a parallelepiped 
be constructed with its edges parallel to the three 
given vectors and with its diagonal equal to the vector whose compo- 
nents are desired. The edges of the parallelepiped are then certain 






164 DIFFERENTIAL CALCULUS 

•multiples xa, yft, zy of a, ft, y, and these are the desired components 
of p. The vector p may be written as 

p = xa + yp + zy* (31) 

It is clear that two equal vectors would necessarily have the same 
components along three given directions and that the components of a 
zero vector would all be zero. Just as the equality of two complex 
numbers involved the two equalities of the respective real and imagi- 
nary parts, so the equality of two vectors as 

p = xa + yft -My = x'a + y'ft + z'y = p' (31') 

involves the three equations x = x 1 , y = y', z = z'. 

As a problem in the use of vectors let there be given the three vectors a, p, y 
from an assumed origin O to three vertices of a parallelogram ; required the vector 
to the other vertex, the vector expressions for the sides and diagonals of the paral- 
lelogram, and the proof of the fact that the diagonals bisect 
each other. Consider the figure. The side AB is, by the 
triangle law, that vector which when added to OA = a 
gives OB = jS, and hence it must be that AB = p — a. 
In like manner AC = y — a. Now OD is the sum of OC 
and CD, and CD = AB ; hence OD = y + p - a . The diag- 
onal AD is the difference of the vectors OD and OA, and 
is therefore y + p — 2 a. The diagonal BC is y — /3. Now the vector from to the 
middle point of BC may be found by adding to OB one half of BC. Hence this 
vector is /3 + i (7 — /3) or | (j3 -f 7) . In like manner the vector to the middle point of 
AD is seen to be a + ± ( 7 + p — 2 a) or \ (7 + /3), which is identical with the former. 
The two middle points therefore coincide and the diagonals bisect each other. 

Let a and ft be any two vectors, \a\ and \ft\ their respective lengths, 
and Z (a, ft) the angle between them. For convenience the vectors may 
be considered to be laid off from the same origin. The product of the 
lengths of the vectors by the cosine of the angle between the vectors 
is called the scalar product, 

scalar product = a* ft = \a\\ft\ cos Z (a, ft), (32) 

of the two vectors and is denoted by placing a dot between the letters. 
This combination, called the scalar product, is a number, not a vector. 
As |/2] cos Z (a, ft) is the projection of ft upon the direction of a, the 
scalar product may be stated to be equal to the product of the length 
of either vector by the length of the projection of the other upon it. 
In particular if either vector were of unit length, the scalar product 
Avould be the projection of the other upon it, with proper regard for 

* The numbers x, y, z are the oblique coordinates of the terminal end of p (if the 
initial end be at the origin) referred to a set of axes which are parallel to a, P, 7 and 
upon which the unit lengths are taken as the lengths of a, P, 7 respectively. 



COMPLEX LUMBERS AND VECTORS 165 

the sign ; and if both vectors are unit vectors, the product is the cosine 
of the angle between them. 

The scalar product, from its definition, is commutative so that a»(3=(3»a. 
Moreover (ma)»(3 = a»(m/3) = m (a* (3), thus allowing a numerical factor 
m to be combined with either factor of the product. Furthermore the 
distributive law 

a*((3 + y) = a*p + a>y or (a + 0).y = oc.y + fi.y (33) 

is satisfied as in the case of numbers. For if a be written as the product 
aa 1 of its length a by a vector a x of unit length in the direction of a, 
the first equation becomes 

aa^ffl + y) = aa^/3 + aa^y or a^{fi + y) = a^ft -f a^y. 

And now a »(f$ + y) is the projection of the sum /? + y upon the direc- 
tion of a, and a^fi + a^y is the sum of the projections of /3 and y upon 
this direction; by the law of projections these are equal and hence the 
distributive law is proved. 

The associative law does not hold for scalar products ; for (a»(3) y 
means that the vector y is multiplied by the number a»(3, whereas 
a ()3«y) means that a is multiplied by (j8»y), a very different matter. 
The laws of cancellation cannot hold ; for if 

a./3 = 0, then | a j [ (3 \ cos Z (a, (3) = 0, (34) 

and the vanishing of the scalar product a»/3 implies either that one of 
the factors is or that the two vectors are perpendicular. In fact 
a»(3 = is called the condition of perpendicularity. It should be noted, 
however, that if a vector p satisfies 

p .a = 0, P .p = 0, p .y = 0, (35) 

three conditions of perpendicularity with three vectors a, /3, y not 
parallel to the same plane, the inference is that p = 0. 
76. Another product of two vectors is the vector product, 

vector product = ax (3 = v \a\ |/3| sin Z (a, (3), (36) 

where v represents a vector of unit length normal to the plane of a 
and (3 upon that side on which rotation from a to 
j3 through an angle of less than 180° appears posi- ox/3' 
tive or counterclockwise. Thus the vector product 
is itself a vector of which the direction is perpen- 
dicular to each factor, and of which the magni- "qS ^ ^^ 
tude is the product of the magnitudes into the 
sine of the included angle. The magnitude is therefore equal to the 
area of the parallelogram of which the vectors a and (3 are the sides. 



166 DIFFERENTIAL CALCULUS 

The vector product will be represented by a cross inserted between the 
letters. 

As rotation from (3 to a is the opposite of that from a to /3, it follows 
from the definition of the vector product that 

fixa = — ax/3, not ax/3 = ^x^ (37) 

and the product is not commutative, the order of the factors must be 
carefully observed. Furthermore the equation 

ax/3 = v\a\\/3\smZ(a, /?) = (38) 

implies either that one of the factors vanishes or that the vectors a and 
f3 are parallel. Indeed the condition ax/3 = is called the condition of 
■parallelism. The laws of cancellation do not hold. The associative law 
also does not hold ; for (ax/3)xy is a vector perpendicular to ax/3 and y, 
and since ax/3 is perpendicular to the plane of a and /3, the vector (ax/3)xy 
perpendicular to it must lie in the plane of a and f3 ; whereas the vec- 
tor <*x(/?xy) ? by similar reasoning, must lie in the plane of /3 and y ; and 
hence the two vectors cannot be equal except in the very special case 
where each was parallel to /3 which is common to the two planes. 

But the operation (ma)x(3 = ax(-m/3) = m(ax(3^, which consists in 
allowing the transference of a numerical factor to any position in the 
product, does hold ; and so does the distributive law 

#x(/3"+ y) = ax/3 + axy and (a + /3)xy = axy -f /3xy, (39) 

the proof of which will be given below. In expanding according to 
the distributive law care must be exercised to keep the order of the 
factors in each vector product the same on both sides of the equation, 
owing to the failure of the commutative law ; an interchange of the 
order of the factors changes the sign. It might seem as if any algebraic 
operations where so many of the laws of elementary algebra fail as in 
the case of vector products would be too restricted to be very useful ; 
that this is not so is due to the astonishingly great number of problems 
in which the analysis can be carried on with only the laws of addition 
and the distributive law of multiplication combined with the possibility 
of transferring a numerical factor from one position to another in a 
product ; in addition to these laws, the scalar product a»(3 is commuta- 
tive and the vector product ax/3 is commutative except for change of sign. 
In addition to segments of lines, plane areas may be regarded as 
vector quantities ; for a plane area has magnitude (the amount of the 
area) and direction (the direction of the normal to its plane). To specify 
on which side of the plane the normal lies, some convention must be 
made. If the area is part of a surface inclosing a portion of space, the 



COMPLEX NUMBERS AND VECTORS 



167 



Ak 




normal is taken as the exterior normal. If the area lies in an isolated 

plane, its positive side is determined only in connection with some 

assigned direction of description of its bounding curve ; the rule is : If 

a person is assumed to walk along the boundary of an area in an 

assigned direction and upon that side of the plane which 

causes the inclosed area to lie upon his left, he is said 

to be upon the positive side (for the assigned direction 

of description of the boundary), and the vector which 

represents the area is the normal to that side. It has 

been mentioned that the vector product represented 

an area. 

That the projection of a plane area upon a given plane gives an area 
which is the original area multiplied by the cosine of the angle between 
the two planes is a fundamental fact of projection, following from the 
simple fact that lines parallel to the intersection of the two planes are 
unchanged in length whereas lines perpendicular to the intersection 
are multiplied by the cosine of the angle between the planes. As the 
angle between the normals is the same as that between the planes, the 
projection of an area upon a plane and the projection of the vector rep- 
resenting the area upon the normal to the plane are equivalent. The 
projection of a closed area upon a plane is zero ; for the area in the 
projection is covered twice (or an even number of times) with opposite 
signs and the total algebraic sum is therefore 0. 

To prove the law ax(/3 -\- y) = ax (3 -+- axy and illustrate the use of 
the vector interpretation of areas, construct a triangular prism with the 
triangle on /?, y, and /3 + y as base and a as lateral edge. The total 
vector expression for the surface of this prism is 

/Sxa + yxa + ax(J3 + y) + K0*y) - h P*y = 0, 

and vanishes because the surface is closed. A cancel- 
lation of the equal and opposite terms (the two 
bases) and a simple transposition combined with the 
rule ftxa = — axfi gives the result 

#x(/3 + y) = — fixa — yxa = axj3 -f- axy. 

A system of vectors of reference which is particularly useful consists 
of three vectors i, j, k of unit length directed along the axes X } Y, Z 
drawn so that rotation from X to Y appears positive from the side of 
the a^-plane upon which Z lies. The components of any vector r drawn 
from the origin to the point (x, y, z) are 

xi, y], z\sl, and r = xi -f y] -f- zk.. 




168 DIFFEEENTIAL CALCULUS 

The products of i, j, k into each other are, from the definitions, 

i.i = j.j = k.k = 1, 
i.j = j.i = j.k = k.j = k.i = i.k = 0, 

lxi == jxj = kxk = 0, v 7 

ixj = — jxi = k, jxk == — kxj = i, kxi = — ixk = j. 

By means of these products and the distributive laws for scalar and 
vector products, any given products may be expanded. Thus if 

a = a x i -f a 2 ] -f a g k and £ = & x i + bj -f 6 3 k, 

then a.fi = afc + aj> 2 + a^, (41) 

ax ^ = (« 2 & 3 - a jj 2 )i + (« 3 ^ 1 - «/> 3 ) j + (a^ - «/^)k, 

by direct multiplication. In this way a passage may be made from 
vector formulas to Cartesian formulas whenever desired. 

EXERCISES 

1 . Prove geometrically that a + (/3 + 7) = (a + ft) + 7 and ra(a + /3) = ma + m(3. 

2. If a and /3 are the vectors from an assumed origin to A and B and if C 
divides AB in the ratio m : n, show that the vector to C is 7 = (71a + m@)/(m + n). 

3. In the parallelogram ABCD show that the line .Si? connecting the vertex to 
the middle point of the opposite side CD is trisected by the diagonal AD and 
trisects it. * 

4 . Show that the medians of. a triangle meet in a point and are trisected. 

5 . If Wj and m 2 are two masses situated at P 1 and P 2 , the center of gravity or 
center of mass of m 1 and ?n 2 is defined as that point G on the line P 1 P 2 which 
divides P X P 2 inversely as the masses. Moreover if G 1 is the center of mass of a 
number of masses of which the total mass is M x and if G 2 is the center of mass of 
a number of other masses whose total mass is M 2 , the same rule applied to M x and 
M 2 and G x and G 2 gives the center of gravity G of the total number of masses. 
Show that 

?n 1 r 1 + m 2 r 2 _ _ m l r 1 -f ra 2 r 4- • • • + wi w r« _ S?/?r 

m 1 + m 2 m 1 + ^ 2 + * " " + m n 2m ' 

where f denotes the vector to the center of gravity. Resolve into components to 

sll0W _ _ 2mx _ _ Xmy _ _ 2mz 

_ 2m ' ~ 2m ' ~ ~~ 2m 

6. If a and /3 are two fixed vectors and p a variable vector, all being laid off 
from the same origin, show that (p — j3)»a = is the equation of a plane through 
the end of /3 perpendicular to a. 

7. Let or, /3, 7 be the vectors to the vertices ^4., 5, C of a triangle. Write the 
three equations of the planes through the vertices perpendicular to the opposite 
sides. Show that the third of these can be derived as a combination of the other 
two ; and hence infer that the three planes have a line in common and that the 
perpendiculars from the vertices of a triangle meet in a point. 



COMPLEX NUMBERS AXD VECTOES 169 

8. Solve the problem analogous to Ex. 7 for the perpendicular bisectors of the 
sides. 

9. Note that the length of a vector is Voxar. If a, /3, and y = /3 — a are the 
three sides of a triangle, expand 7.7 = {0— a).(/3 — a) to obtain the law of cosines. 

10. Show that the sum of the squares of the diagonals of a parallelogram equals 
the sum of the squares of the sides. What does the difference of the squares of the 
diagonals equal ? 

11. Show that - — - a and - — — — are the components of /3 parallel and perpen- 

dicular to a by showing 1° that these vectors have the right direction, and 2° that 
they have the right magnitude. 

12. If a, /3, 7 are the three edges of a parallelepiped which start from the same 
vertex, show that (ax/3). 7 is the volume of the parallelepiped, the volume being 
considered positive if 7 lies on the same side of the plane of a and /3 with the 
vector ax/3. 

13. Show by Ex. 12 that (ax/3).y = a»(pxy) and (ax@)»y = (j3xy).a; and hence 
infer that in a product of three vectors with cross and dot, the position of the cross 
and dot may be interchanged and the order of the factors may be permuted cyc- 
lically without altering the value. Show that the vanishing of (ax/3). 7 or any of 
its equivalent expressions denotes that a, /3, 7 are parallel to the same plane ; the 
condition ax/3.7 = is called the condition of complanarity. 

14. Assuming a = a x i + a. 2 ) + a 3 k, (3 — &ji -f 6 2 j + 6 3 k, 7 = c a i + c. 2 j + c 3 k, 
expand a.7, a«/3, and ax(pxy) in terms of the coefficients to show 

ax(/3x7) = (a»7)|8 — (a*(3)y; and hence (ax/3)x7 = (a»y) j3 — (yp) a. 

15. The formulas of Ex. II for expanding a product with two crosses and the 
rule of Ex. 13 that a dot and a cross may be interchanged may be applied to expand 

(ax/3)x(7x5) = (a»y*8)p— ((S*yx8)a = (ax/3.5) 7— (axp»y)8 
and (ax/3). ( 7 x5) = (a. 7 )(/3-5) - (/3. 7 )(a.5). 

16. If a and /3 are two unit vectors in the z?/-plane inclined at angles and <f> 
to the ic-axis, show that 

a = i cos 6 + j sin 0, /3 = i cos + j sin ; 

and from the fact that a«/3 = cos(0 — 6) and ax/3 = ksin(0 — 6) obtain by multi- 
plication the trigonometric formulas for sin (0 — 6) and cos (0 — 6) . 

17. If I, m, n are direction cosines, the vector li + mj + nk is a vector of unit 
length in the direction for which Z, m, n are direction cosines. Show that the 
condition for perpendicularity of two directions (I, m, n) and (V, to', n') is 
IV + mm" + nn' = 0. 

18. With the same notations as in Ex. II show that 
a* a = a'l + a'l + a 3 2 and ax ft 



i j k 






a x a 2 a z 


a x a 2 a s 


and 


axp.y = 


h \ \ 


\ 6 2 6 3 






e t c 2 c 3 



170 



DIFFEEE^TIAL CALCULUS 



19. Compute the scalar and vector products of these pairs of vectors : 
f 6i + 0.3 j- 5k 



(a) |o.li-4.2j + 2.5k, (/3) |_3i-2j + k, 



(7) 



/i + k 
U + i. 



20. Find the areas of the parallelograms defined by the pairs of vectors in 
Ex. 19. Find also the sine and cosine of the angles between the vectors. 

21. Prove ax[(3x(yx8)~\ = (a.yx5)l3 — a>/3yx5 — (3-5 axy — /3.y ax5. 

22. What is the area of the triangle (1, 1, 1), (0, 2, 3), (0, 0, - 1) ? 

77. Vector differentiation. As the fundamental rules of differentia- 
tion depend on the laws of subtraction, multiplication by a number, 
the distributive law, and the rules permitting rearrangement, it follows 
that the rules must be applicable to expressions containing vectors 
without any changes except those implied by the fact that axfi =?= fixa. 
As an illustration consider the application of the definition of differen- 
tiation to the vector product Uxv of two vectors which are supposed 
to be functions of a numerical variable, say x. Then 

A(UxV) = (U + Au)x(v + AV) — UxV 
= UxAv + AUxv + AUxAv, 

A(Uxv) Av Au , AUxAv 
-^ L = Ux— + —xV H , 

Ax Ax Ax Ax 



f/(uxv) ... A(uxv) dv 

— — — - = Inn — = ux— - -+- — 



dx 



Ax 



dx 



du 

dx 



Here the ordinary rule for a product is seen to hold, except that 
the order of the factors must not be interchanged. 

The interpretation of the derivative is important. Let the variable 
vector r be regarded as a function of some variable, say x, and suppose 
r is laid off from an assumed origin so that, as x varies, 
the terminal point of r describes a curve. The incre- 
ment Ar of r corresponding to. Ax is a vector quantity 
and in fact is the chord of the curve as indicated. 
The derivative 



dx ,. Ar 
— = Inn — : 

dx Ax 



dx Ar 

— = Inn — = t 
as As 



(42) 




is therefore a, vector tangent to the curve; in particular if 
the variable x were the arc s, the derivative would have 
the magnitude unity and would be a unit vector tangent to the curve. 
The derivative or differential of a vector of constant length is per- 
pendicular to the vector. This follows from the fact that the vector 



COMPLEX NUMBERS AND VECTORS 171 

then describes a circle concentric with the origin. It may also be seen 
analytically from the equation 

d(r.r) = dr.r + iv/r = 2 iv/r = d const. = 0. (43) 

If the vector of constant length is of length unity, the increment Ar is 
the chord in a unit circle and, apart from infinitesimals of higher 
order, it is equal in magnitude to the angle subtended at the center. 
Consider then the derivative of the unit tangent t to a curve with 
respect to the arc s. The magnitude of dt is the angle the tangent turns 
through and the direction of dt is normal to t and hence to the curve. 
The vector quantity, ^ 7^ 

curvature C = -7- = -7-5 > (44) 

ds ds 2 v y 

therefore has the magnitude of the curvature (by the definition in § 42) 
and the direction of the interior normal to the curve. » 

This work holds equally for plane or space curves. In the case of a space curve 
the plane which contains the tangent t and the curvature C is called the osculating 
plane (§ 41). By definition (§ 42) the torsion of a space curve is the rate of turning 
of the osculating plane with the arc, that is, d\f//ds. To find the torsion by vector 
methods let c be a unit vector C/Vc»C along C. Then as t and c are perpendicular, 
n = txc is a unit vector perpendicular to the osculating plane and dn will equal df 
in magnitude. Hence as a vector quantity the torsion is 

_ dn d (txc) dt , , dc ..dc 

T = — = — = — xc + tx— = tx— , (45) 

ds ds ds ds ds 

where (since dt/ds = C, and c is parallel to C) the first term ^ 

drops out. Next note that dn is perpendicular to n because it 

is the differential of a unit vector, and is perpendicular to t 

because dn = cZ(txc) = txdc and t.(txdc) = since t, t, dc are 

necessarily complanar (Ex. 12, p. 169). Hence T is parallel 

to c. It is convenient to consider the torsion as positive when n 

the osculating plane seems to turn in the positive direction when 

viewed from the side of the normal plane upon which t lies. An inspection of the 

figure shows that in this case dn has the direction — c and not + c. As c is a unit 

vector, the numerical value of the torsion is therefore — c»T. Then 

T= — ■ dc . d C 





= - ctx — + C = — ctx (450 

** Vcc 

. C dH 

= t x 

C-C cfe 3 r".r" 
where differentiation with respect to s is denoted by accents. 

78. Another sort of relation between vectors and differentiation 
comes to light in connection with the normal and directional deriva- 
tives (§ 48). If F(x, y, z) is a function which has a definite value at 




172 DIFFERENTIAL CALCULUS 

each point' of space and if the two neighboring surfaces F = C and 
F = C + dC are considered, the normal derivative of F is the rate of 
change of F along the normal to the surfaces and 
is written dF/dn. The rate of change of F along 
the normal to the surface F = C is more rapid than 
along any other direction ; for the change in F be- 
tween the two surfaces is dF = dC and is constant, 
whereas the distance dn between the two surfaces is 
least (apart from infinitesimals of higher order) along the normal. In 
fact if dr denote the distance along any other direction, the relations 
shown by the figure are 

dr = sec Odn and — — = — — cos 0. (46) 

dr dn 

If now n denote a vector of unit length normal to the surface, the 
product ndF/dnwill be a vector quantity which has both the magnitude 
and the direction of most rapid increase of F. Let 

dF 

n ^n~ = VF=gmdF (47) 

be the symbolic expressions for this vector, where VF is read as "del F" 

and grad F is read as " the gradient of F." If dr be the vector of which 

dr is the length, the scalar product n»dr is precisely cos 6dr, and hence 

it follows that 

dF 
dr.VF=dF and r^VF = — : , (48) 

where r 1 is a unit vector in the direction dr. The second of the equa- 
tions shows that the directional derivative in any direction is the com- 
ponent or projection of the gradient in that direction. 

From this fact the expression of the gradient may be found in terms 
of its components along the axes. For the derivatives of F along the 
axes are dF/dx, dF/dy, dF/dz, and as these are the components of VF 
along the directions i, j, k, the result is 

.dF .dF , dF 
VF = grad i^l — -fl-^--fk^-- 

a % \ W 

Hence V = i — -f-j — + k:r- 

cx cy cz 

may be regarded as a symbolic vector-differentiating operator which 
when applied to F gives the gradient of F. The product 

dt.WF = (dx j- x + dy~ + dz £\F= dF (50) 



COMPLEX NUMBERS AND VECTORS 173 

is immediately seen to give the ordinary expression for dF. From this 
form of grad F it does not appear that the gradient of a function is 
independent of the choice of axes, but from the manner of derivation 
of VF first given it does appear that grad F is a definite vector quan- 
tity independent of the choice of axes. 

In the case of any given function F the gradient may be found by 
the application of the formula (49) ; but in many instances it may also 
be found by means of the important relation dx»VF = dF of (48). For 
instance to prove the formula V (FG) = FVG -f- GVF, the relation may 
be applied as follows : 

dr.V(FG) = d(FG) = FdG + GdF 

= Fdr.VG + Gdx.VF = dx-(FVG + GVF). 

Now as these equations hold for any direction dr, the dx may be can- 
celed by (35), p. 165, and the desired result is obtained. 

The use of vector notations for treating assigned practical problems involving 
computation is not great, but for handling the general theory of such parts of 
physics as are essentially concerned with direct quantities, mechanics, hydro- 
mechanics, electromagnetic theories, etc., the actual use of the vector algorisms 
considerably shortens the formulas and has the added advantage of operating 
directly upon the magnitudes involved. At this point some of the elements of 
mechanics will be developed. 

79. According to Newton's Second Law, when a force acts upon a 
particle of mass m, the rate of change of momentum is equal to the 
force acting, and takes place in the direction of the force. It therefore 
appears that the rate of change of momentum and momentum itself 
are to be regarded as vector or directed magnitudes in the application 
of the Second Law. Now if the vector r, laid off from a fixed origin 
to the point at which the moving mass m is situated at any instant of 
time t, be differentiated with respect to the time t, the derivative dx/dt 
is a vector, tangent to the curve in which the particle is moving and of 
magnitude equal to ds/dt or v, the velocity of motion. As vectors*, 
then, the velocity v and the momentum and the force may be written as 



dx 
v = -, mv, 


F = |(»v). 


dv d 2 x 
dt dt 2 


dv dh 
dt~~ d? 



Hence 

From the equations it appears that the force F is the product of the 
mass m by a vector f which is the rate of change of the velocity regarded 

* In applications, it is visual to denote vectors by heavy type and to denote the magni- 
tudes of those vectors by corresponding italic letters. 



174 DIFFERENTIAL CALCULUS 

as a vector. The vector f is called the acceleration; it must not be con- 
fused with the rate of change dv/dt or d 2 s/dt 2 of the speed or magnitude 
of the velocity. The components f x , f y , f z of the acceleration along the 
axes are the projections of f along the directions i, j, k and may be 
written as f«i, f »j, f »k. Then by the laws of differentiation it follows 

that 

/■ fi_ rfv i_ j ( v>1 ) ^ 

f r = I»l — "T - • 1 — : = — — j 

. x . ^ 2 r . ^ 2 (r.i) d 2 x 
or / x = f .1 = — -1 = — V^ = — 



dt 2 dt 2 dt 2 

d 2 x dhj d 2 z 

Hence f T = — -5 > /,, = — % 5 f = -—, 

Jx dt 2 Jy dt 2 Jz dt 2 

and it is seen that the components of the acceleration are the acceler- 
ations of the components. If X, Y, Z are the components of the force, 
the equations of motion in rectangular coordinates are 

c ^ x ^ dhi ^ d 2 z ' ^^ 

Instead of resolving the acceleration, force, and displacement along 
the axes, it may be convenient to resolve them along the tangent and 
normal to the curve. The velocity v may be written as vt, where v is 
the magnitude of the velocity and t is a unit vector tangent to the 
curve. Then 



But ^ = X^ = C« = 7jn, (53) 



f 


_dv _ 

~~ dt ~ 


d(vt) 

dt 


dv J 


dt 

v — • 
dt 






dt _ 

dt ~~ 


dt ds 
ds dt 




n, 




ius 


of curvature and n is a 


unit 


non 


cl 2 s 
~df 2 


v 2 
t+— n 


1 ft ~ 


d 2 s 

= dt 2 ' 


J n 


_v 2 
" R 



(53') 

It therefore is seen that the component of the acceleration along the 
tangent is d 2 s/dt 2 , or the rate of change of the velocity regarded as a 
number, and the component normal to the curve is v 2 /R. If T and N 
are the components of the force along the tangent and normal to the 
curve of motion, the equations are 

d 2 s - v 2 

T = ™f t = m dtf> N = mf n = m-- 

It is noteworthy that the force must lie in the osculating plane. 

If r and r + Ar are two positions of the radius vector, the area of 
the sector included by them is (except for infinitesimals of higher order) 



COMPLEX NUMBEES AXD VECTOKS 175 

AA = i-rx(r + Ar) = JrxAr, and is a vector quantity of which the 
direction is normal to the plane of r and r + Ar, that is, to the plane 
through the origin tangent to the curve. The rate of description of area, 
or the a real velocity, is therefore 

dk ' , Ar . dx . ,- 

_ == l lmi rx-=>rx-= i rxv. (54) 

The projections of the areal velocities on the coordinate planes, which 
are the same as the areal velocities of the projection of the motion on 
those planes, are (Ex. 11 below) 

2{ V dt ~ dt)' 2\~ dt X dt)' 2\ X dt ' J dt)- {oi > 

If the force F acting on the mass m passes through the origin, then 
r and F lie along the same direction and rxF = 0. The equation of 
motion may then be integrated at sight. 

dv „ dv - _ n 

m — = F, mrx— = rxF = 0, 

dt ' dt 

dv d 

rx— = — (rxv) = 0, rxv = const. 

It is seen that in this case the rate of description of area is a constant 
vector, which means that the rate is not only constant in magnitude 
but is constant in direction, that is, the path of the particle m must lie 
in a plane through the origin. When the force passes through a fixed 
point, as in this case, the force is said to be central. Therefore when a 
particle moves under the action of a central force, the motion takes place 
in a plane passing through the center and the rate of description of 
areas, or the areal velocity, is constant. 

80. If there are several particles, say ?i, in motion, each has its own equation 
of motion. These equations may be combined by addition and subsequent reduction. 

d 2 !, d 2 r 9 d 2 x n _ 

m-, — ± = F, , m 9 — f = F 9 , • • •, m n — - 1 = F n , 

1 dt 1 1 ' 2 dt 2 *' ' dt 2 

d 2 !-, d 2 r d 2 r„ „ „ „ 

and m 1 — 1 + m 2 — f + . . . + m n — f = F x + F 9 + • ■ • + F„. 

dt 2 dt 2 dt 2 

d 2 i d 2 x d 2 r d 2 

But m t — ' + m, -J + ■ ■ • + *. — » = - (m^ + ,» 2 r 2 + . . . + m„r„). 

Let ??i 1 r 1 + m 2 T 2 + • • • + w^r,, = (m 1 + m 2 + • ■ • + m n ) r = M f 



or 



L r T + m 2 r 2 + • • • + m n r n _ Smr _ 2mr 
m 1 + m 2 + • • • + m n 2m M 

.d 2 r 



^ = Fi + F 2 + • ■ • + F n = 2)F. (55) 



176 DIFFERENTIAL CALCULUS 

Now the vector r which has been here introduced is the vector of the center of 
mass or center of gravity of the particles (Ex. 5, p. 168). The result (55) states, on 
comparison with (51), that the center of gravity of the n masses moves as if all the 
mass M were concentrated at it and all the forces applied there. 

The force F; acting on the ith mass may be wholly or partly due to attractions, 
repulsions, pressures, or other actions exerted on that mass by one or more of the 
other masses of the system of n particles. In fact let F t - be written as 

Fi = F l0 + F a + F i2 + • • • + F iM , 

where F# is the force exerted on m t - by m,j and F t - is the force due to some agency 
external to the n masses which form the system. Now by Newton's Third Law, 
when one particle acts upon a second, the second reacts upon the first with a 
force which is equal in magnitude and opposite in direction. Hence to Fy above 
there will correspond a force Fji = — F# exerted by m ? - on rrij. In the sum 2F; all 
these equal and opposite actions and reactions will drop out and 2F; may be re- 
placed by SFjo, the sum of the external forces. Hence the important theorem that : 
The motion of the center of rnass of a set of particles is as if all the mass were concen- 
trated there and all the external forces were applied there (the internal forces, that is, 
the forces of mutual action and reaction between the particles being entirely 
neglected). 

The moment of a force about a given point is defined as the product of the force 
by the perpendicular distance of the force from the point. If r is the vector from 
the point as origin to any point in the line of the force, the moment is therefore 
rxF when considered as a vector quantity, and is perpendicular to the plane of the 
line of the force and the origin. The equations of n moving masses may now be 
combined in a different way and reduced. Multiply the equations by r x , r 2 , • • •, i n 
and add. Then 

dv, dv dv n _ _ „ 

m i r i X "77 + m 2 r 2 X ~rr + •■■ + ™n*n* -77 = r i xF l + r 2 xF 2 + / ' * + r « xF « 
(XL (XL (XL 

d d d _ _. . „ 

or m-, — r, xv, + m 9 — r„xy 9 + • • • + m n — r K xv„ = r, xF-. + r„xF„ + • • • + r„xF M 
dt dt dt 

or — (ra 1 r 1 xv 1 + m 2 r 2 xv 2 + • • • + m„r M xv„) = SrxF. " (56) 

This equation shows that if the areal velocities of the different masses are multiplied 
by those masses, and all added together, the derivative of the sum obtained is equal 
to the moment of all the forces about the origin, the moments of the different forces 
being added as vector quantities. 

This result may be simplified and put in a different form. Consider again the 
resolution of F 4 - into the sum F l0 + F,i + • • • + F,„, and in particular consider the 
action Fy and the reaction F,- t = — F# between two particles. Let it be assumed 
that the action and reaction are not only equal and opposite, but lie along the line 
connecting the two particles. Then the perpendicular distances from the origin to 
the action and reaction are equal and the moments of the action and reaction are 
equal and opposite, and may be dropped from the sum 2r ? -xF;, which then reduces 
to 2r;xFjo. On the other hand a term like miti*Vi may be written as rix(m t -v t -). This 
product is formed from the momentum in exactly the same way that the moment 
is formed from the force, and it is called the moment of momentum. Hence the 
equation (56) becomes 



COMPLEX NUMBERS AXD VECTORS 177 

— (total moment of momentum) = moment of external forces. 

Hence the result that, as vector quantities : The rate of change of the moment of 
momentum of a system of particles is equal to the moment of the external forces (the 
forces between the masses being entirely neglected under the assumption that action 
and reaction lie along the line connecting the masses). 

EXERCISES 

1. Apply the definition of differentiation to prove 

(a) d (u.v) = U'dv + v.du, (/3) d [u.(vxw)] = du.(vxw) + u.(cZvxw) + u.(vxcZw). 

2. Differentiate under the assumption that vectors denoted by early letters of 
the alphabet are constant and those designated by the later letters are variable : 

(a) ux(vxw), (/3) acosi + bsint, (7) (u.u)u, 

/*N dU I X / dU Cl ' 2U \ /v\ / 

3. Applv the rules for change of variable to show that — = — ■ ■ — , where 

11 J ° ds 2 s' 3 

acce nts d enote differentiation with respect to x. In case r = xi -f y] show that 
1/ VC«C takes the usual form for the radius of curvature of a plane curve. 



4. The equation of the helix is r = ia cos + ja sin + kb<p with s = Va 2 + h' 2 ; 
show that the radius of curvature is (a 2 + k 2 )/a. 

5. Find the torsion of the helix. It is b/(a 2 + k 2 ). 

6. Change the variable from s to some other variable t in the formula for torsion. 

7. In the following cases find the gradient either by applying the formula which 
contains the partial derivatives, or by using the relation dr» VF — dF, or both : 

(a) r.r = x 2 + y 2 + z 2 , (/3) log r, (7) r = VrTr", 
(5) log(x2 + y 2 ) = log [r.r - (k-r)*j, (e) (rxa).(rxb). 

8. Prove these laws of operation with the symbol V : 

( a ) V(F+ G) = VF+ VG, 0) G 2 V(F/G) = GYF- FVG. 

9. If r, are polar coordinates in a plane and ^ is a unit vector along the radius 
vector, show that dr \/dt = nd<p/dt where n is a unit vector perpendicular to the 
radius. Thus differentiate r = rr l twice and separate the result into components 
along the radius vector and perpendicular to it so that 

d 2 r /rZ0\' 2 . d 2 4> , n d<j> dr ld/ 9 d<f> 

f r = r I — I , f& = r —^ + 2 — - — = ( r 2 — 

dt 2 \dt) * dt 2 dt dt rdt\ dt 

10. Prove conversely to the text that if the vector rate of description of area is 
constant, the force must be central, that is, rxF = 0. 

11. Note that rxvi, rxyj, rxv«k are the projections of the areal velocities upon 
the planes x = 0, y = 0, z = 0. Hence derive (54') of the text. 



178 DIFFEEENTIAL CALCULUS 

12. Show that the Cartesian expressions for the magnitude of the velocity and 
of the acceleration and for the rate of change of the speed dv/dt are 

x'x" + y'y" + z'z" 



Vx' 2 + y' 2 + z' 2 , / = Vx" 2 + y" 2 + z" 2 



Vx' 2 + y' 2 + z' 2 
where accents denote differentiation with respect to the time. 

13. Suppose that a body which is rigid is rotating about an axis with the 
angular velocity w = d<p/dt. Represent the angular velocity by a vector a drawn 
along the axis and of magnitude equal to w. Show that the velocity of any point 
in space is v = axr, where r is the vector drawn to that point from any point of 
the axis as origin. Show that the acceleration of the point determined by r is in a 
plane through the point and perpendicular to the axis, and that the components are 

ax (axr) = (a«r)a — w 2 r toward the axis, (da/cW)xr perpendicular to the axis, 

under the assumption that the axis of rotation is invariable. 

14 . Let r denote the center of gravity of a system of particles and r/ denote the 
vector drawn from the center of gravity to the ith particle so that r 4 - = r + r/ and 
W-= v + v/. The kinetic energy of the ith particle is by definition 

|m,-D- = |miVi»V{ = \mi(y + v/).(v + v/). 

Sum up for all particles and simplify by using the fact 2m r -r^ = 0, which is due to 
the assumption that the origin for the vectors r- is at the center of gravity. Hence 
prove the important theorem : The total kinetic energy of a system is equal to the 
kinetic energy which the total mass would have if moving with the center of gravity 
plus the energy computed from the motion relative to the center of gravity as origin, 

that is, 

T = \ ZmiV? = i M v 2 + 1 Sm^ 2 . 

15. Consider a rigid body moving in a plane, which may be taken as the xy- 
plane. Let any point r of the body be marked and other points be denoted rela- 
tive to it by t. The motion of any point r' is compounded from the motion of r 
and from the angular velocity a = ko> of the body about the point r . In fact the 
velocity v of any point is v == v + axr. Show that the velocity of the point denoted 
by r' = kxv /w is zero. This point is known as the instantaneous center of rotation 
(§ 39). Show that the coordinates of the instantaneous center referred to axes at 
the origin of the vectors r are 

1 dy n 1 dx 

x = r,i = x - --£-°, y = r.j = y ■+. - -f- 
u dt w dt 

16. If several forces F x , F 2 , • • •, F n act on a body, the sum R = 2Fj is called 
the resultant and the sum S^xF*, where x { is drawn from an origin to a point 
in the line of the force F{, is called the resultant moment about 0. Show that the 
resultant moments M and M / about two points are connected by the relation 
M 0/ = Mo + M /(R ), where Mo'(R<y) means the moment about 0' of the resultant 
R considered as applied at 0. Infer that moments about all points of any line 
parallel to the resultant are equal. Show that in any plane perpendicular to R 
there is a point 0' given by r = RxM /R-R, where is any point of the plane, 
such that Mo^ is parallel to R. 



PAET II. DIFFEEEXTIAL EQUATIOXS 

CHAPTER VII 

GENERAL INTRODUCTION TO DIFFERENTIAL EQUATIONS 

81. Some geometric problems. The application of the differential 
calculus to plane curves has given a means of determining some 
geometric properties of the curves. For instance, the length of the 
subnormal of a curve (§7) is ydy/dx, which in the case of the parabola 
y 2 = ±px is 2 p. that is, the subnormal is constant. Suppose now it 
were desired conversely to find all curves for which the subnormal is 
a given constant m. The statement of this problem is evidently con- 
tained in the equation 

y j = m or yy = m or ydy = max. 

Again, the radius of curvature of the lemniscate r 2 = a 2 cos 2 <£ is found 
to be it = a 2 /3 r, that is, the radius of curvature varies inversely as the 
radius. If conversely it were desired to find all curves for which the 
radius of curvature varies inversely as the radius of the curve, the state- 
ment of the problem would be the equation 



4- 



m' _ 



d 2 r n /dr 

d<f> 2 ^ [defy 

where k is a constant called a factor of proportionality.* 

Equations like these are unlike ordinary algebraic equations because, 
in addition to the variables x, y or r, <$> and certain constants m or k, 
they contain also derivatives, as dy/dx or dr/dcf> and d 2 r/d<$> 2 . of one of 
the variables with respect to the other. An equation which contains 

* Many problems in geometry, mechanics, and physics are stated in terms of varia- 
tion. For purposes of analysis the statement x varies as y, or x ac y, is written as :c = ky, 
introducing a constant k called a factor of proportionality to convert the variation into 
an equation. In like manner the statement x varies inversely as y, or x & 1/y. becomes 
x = k/y, and a: varies jointly with y and z becomes x — kyz. 

179 



180 DIFFERENTIAL EQUATIONS 

derivatives is called a differential equation. The order of the differential 
equation is the order of the highest derivative it contains. The equa- 
tions above are respectively of the first and second orders. A differen- 
tial equation of the first order may be symbolized as $ (x, y, y') = 0, 
and one of the second order as $(x, y, y r , y") = 0. A function y = f(x) 
given explicitly- or defined implicitly by the relation F(x, y) = is 
said to be a solution of a given differential equation if the equation is 
true for all values of the independent variable x when the expressions 
for y and its derivatives are substituted in the equation. 

Thus to show that (no matter what the value of a is) the relation 
4 ay — x 2 + 2 a 2 log x = 
gives a solution of the differential equation of the second order 



1 + 



\dx) \dx 2 / 



it is merely necessary to form the derivatives 

dx x dx 2 x 2 

and substitute them in the given equation together with y to see that 

W \dx 2 ) 4a 2 \ x 2 ) 4a 2 \ x 2 xV 

is clearly satisfied for all values of x. It appears therefore that the given relation 
for y is a solution of the given equation. 

To integrate or solve a differential equation is to find all the functions 
which satisfy the equation. Geometrically speaking, it is to find all the 
curves which have the property expressed by the equation. In mechan- 
ics it is to find all possible motions arising from the given forces. The 
method of integrating or solving a differential equation depends largely 
upon the ingenuity of the solver. In many cases, however, some method 
is immediately obvious. For instance if it be possible to separate the 
variables, so that the differential dy is multiplied by a function of y 
alone and dx by a function of x alone, as in the equation 

<f> (y) dy = xp (x) dx, then I <f> (y) dy = I if/ (x) dx + C (1) 

will clearly be the integral or solution of the differential equation. 

As an example, let the curves of constant subnormal be determined. Here 

ydy = mdx and y 2 = 2 mx + C. 

The variables are already separated and the integration is immediate. The curves 
are parabolas with semi-latus rectum equal to the constant and with the axis 



GENERAL INTRODUCTION 181 

coincident with the axis of x. If in particular it were desired to determine that 
curve whose subnormal was m and which passed through the origin, it would 
merely be necessary to substitute (0, 0) in the equation y 2 = 2 mx + C to ascertain 
what particular value must be assigned to C in order that the curve pass through 
(0. 0). The value is C = 0. 

Another example might be to determine the curves for which the ^-intercept 
varies as the abscissa of the point of tangency. As the expression (§ 7) for the 
z-intercept is x — ydx/dy, the statement is 

dx . /i' i\ d x 

x — y — = kx or (1 — k)x = y 

dy dy 

Hence (1 - k) d l = c l^ and (1 - k) los y = log x + C. 

y x 

If desired, this expression may be changed to another form by using each side of 
the equality as an exponent with the base e. Then 

efl-fylogy = e logx+C or yl-l- — e C x — Q' x% 

As Cis an arbitrary constant, the constant C = e c is also arbitrary and the solution 
may simply be written as y 1 - k — Cx, where the accent has been omitted from the 
constant. If it were desired to pick out that particular curve which passed through 
the point (1, 1), it would merely be necessary to determine C from the equation 

li-* = CI, and hence (7 = 1. 

As a third example let the curves whose t angent is constant and equal to a be 
determined. The length of the tangent is y Vl + y'-/y' and hence the equation is 



Vl + if 1 ., 1 + y' 2 ., Va 2 — y 2 , 
y -!-£- = a or y 2 J = a or 1 = — y' 

V' y' 2 V 

The variables are therefore separable and the results are 



dx = — dy and x + C = V a 2 — y 2 — a log — ■ 

y y 

If it be desired that the tangent at the origin be vertical so that the curve passes 
through (0, a), the constant C is 0. The curve is the tractrix or '■ curve of pursuit" 
as described by a calf dragged at the end of a rope by a person walking along 
a straight line. 

82. Problems which involve the radius of curvature will lead to differ- 
ential equations of the second order. The method of solving such 
problems is to reduce the equation, if possible, to one of the first order. 
For the second derivative may be written as 

dx y dy 



182 DIFFEEENTIAL EQUATIONS 

is the expression for the radius of curvature. If it be given that the 
radius of curvature is of the form f(x) <f> (?/') ot f(y) <$> (y'), 

(i y ='m+(rt or <U^P!=/o,) 4>(l/'), (3) 

dx dy 

the variables x and y' or y and y' are immediately separable, and an 
integration may be performed. This will lead to an equation of the 
first order ; and if the variables are again separable, the solution may 
be completed by the methods of the above examples. 

In the first place consider curves whose radius of curvature is constant. Then 

" (1 + ?/' 2 )2 dy' dx y' X — C 

— — — a or = — and — = , 

&tf_ (i + ? /2)| a Vl + y' 2 a 
dx 

where the constant of integration has been written as — C/a for future conven- 
ience. The equation may now be solved for y' and the variables become separated 
with the results 

x - C , (x-C) . 

y = — or dy = — = = = ^ = r dx. 

Va 2 ~(x- C) 2 Va 2 - (x - C) 2 

Hence y- C = - Va 2 - (x - C) 2 or (x - C) 2 + (y - C') 2 = a 2 . 

The curves, as should be anticipated, are circles of radius a and with any arbi- 
trary point (C, C) as center. It should be noted that, as the solution has required 
two successive integrations, there are two arbitrary constants C and C" of integra- 
tion in the result. 

As a second example consider the curves whose radius of curvature is double 
the normal. As the length of the normal is y Vl + y" 2 j the equation becomes 

fi±l2i=2»Vr+ipS or i±£ = ±»K 

dy dy 

where the double sign has been introduced when the radical is removed by cancel- 
lation. This is necessary ; for before the cancellation the signs were ambiguous 
and there is no reason to assume that the ambiguity disappears. In fact, if the 
curve is concave up, the second derivative is positive and the radius of curvature 
is reckoned as positive, whereas the normal is positive or negative according as 
the curve is above or below the axis of x ; similarly, if the curve is concave down. 
Let the negative sign be chosen. This corresponds to a curve above the axis and 
concave down, or below the axis and concave up, that is, the normal and the radius 
of curvature have the same direction. Then 

dy = _2y^_ and log?7 = _ log(1 + 2/ , 2) + log2C , 

v i + y 2 

where the constant has been given the form log 2 C for convenience. This expres- 
sion may be thrown into algebraic form by exponentiation, solved for y', and then 



G-EXEKAL IXTKODUCTIOX 183 

y(l + y«) = 20 or ^ = ^ZV or ydy = cZx. 

y V2 CV - y2 

Hence x- C'=C vers- 1 - - V2 Cty - y 2 . 

The curves are cycloids of which the generating circle has an arbitrary radius C 
and of which the cusps are upon the x-axis at the points C ± 2 kirC. If the posi- 
tive sign had been taken in the equation, the curves would have been entirely 
different ; see Ex. 5 (a). 

The number of arbitrary constants of integration which, enter into 
the solution of a differential equation depends on the number of inte- 
grations which are performed and is equal to the order of the equation. 
This results in giving a family of curves, dependent on one or more 
parameters, as the solution of the equation. To pick out any particular 
member of the family, additional conditions must be given. Thus, if 
there is only one constant of integration, the curve may be required 
to pass through a given point ; if there are two constants, the curve 
may be required to pass through a given point and have a given slope 
at that point, or to pass through two given points. These additional 
conditions are called initial conditions. In mechanics the initial condi- 
tions are very important ; for the point reached by a particle describing 
a curve under the action of assigned forces depends not only on the 
forces, but on the point at which the particle started and the velocity 
with which it. started. In all cases the distinction between the constants 
of integration and the given constants of the problem (in the foregoing 
examples, the distinction between C, C and m, k, a) should be kept 
clearly in mind 

EXERCISES 

1 . Verify the solutions of the differential equations : 

(a) xy + \x 2 =C, y'+x + xy' = 0, (/3) x s y 3 (3 e*+ C) = 1, xy'+y+x 4 y±e*=0, 

(7) (1 + x 2 ) tf*=l, 2x= Cev- C^er-V, (5) y + xy* = xV 2 , xy = C 2 x + C, 
(e) V" + V/x = 0, y = G log x + C v (D V = Ce* + G^\ y"+2y = S y', 

,.-... , „ ly/ ^ XVS „ . X VS\ 

(v) V — V — ^ 2 , y — Ce x + e~ 2 x [ C x cos \- C 2 sm ) — x 2 . 

2. Determine the curves which have the following properties: 

(a) The subtangent is constant ; y = Cx m . If through (2, 2), y — 2 1 ~ m x m . 

(j8) The right triangle formed by the tangent, subtangent, and ordinate has the 
constant area k/2 ; the hyperbolas xy + Cy + k = 0. Show that if the curve passes 
through (1, 2) and (2, 1), the arbitrary constant C is and the given k is — 2. 

(7) The normal is constant in length ; the circles (x — C)' 2 + y' 2 = k 2 . 

( 5 ) The normal varies as the square of the ordinate ; catenaries ky = cosh k(x— C) . 
If in particular the curve is perpendicular to the y-axis, C = 0. 

( e ) The area of the right triangle formed by the tangent, normal, and x-axis is 
inversely proportional to the slope ; the circles (x — C)' 2 + y 2 — k. 



184 DIFFERENTIAL EQUATIONS 

3. Determine the curves which have the following properties: 

(a) The angle between the radius vector and tangent is constant ; spirals 
r = Ce k b. 

(|8) The angle between the radius vector and tangent is half that between the 
radius and initial line ; cardioicls r = C(l — cos 0). 

(7) The perpendicular from the pole to a tangent is constant ; r cos (0 — C) = k. 

(5) The tangent is equally inclined to the radius vector and to the initial line ; 
the two sets of parabolas r — C/(l ± cos 0). 

(e) The radius is equally inclined to the normal and to the initial line ; circles 
r = C cos or lines r cos = C. 

4. The arc s of a curve is proportional to the area A, where in rectangular 
coordinates A is the area under the curve and in polar coordinates it is the area 
included by the curve and the radius vectors. From the equation ds = dA show 
that the curves which satisfy the condition are catenaries for rectangular coordi- 
nates and lines for polar coordinates. 

5. Determine the curves for which the radius of curvature 

(a) is twice the normal and oppositely directed ; parabolas (x — C) 2 = C'(2y — C). 

(/3) is equal to the normal and in same direction ; circles (x — C) 2 + y 2 = C' 2 . 

(7) is equal to the normal and in opposite direction ; catenaries. 

(5) varies as the cube of the normal ; conies kCy 2 — C 2 (x + C') 2 = k. 

(e) projected on the z-axis equals the abscissa ; circles. 

( f ) projected on the x-axis is the negative of the abscissa ; catenaries. 
(7?) projected on the cc-axis is twice the abscissa ; central conies. 

(0) is proportional to the slope of the tangent or of the normal. 

83. Problems in mechanics and physics. In many physical problems 
the statement involves an equation between the rate of change of some 
quantity and the value of that quantity. In this way the solution of 
the problem is made to depend on the integration of a differential equa- 
tion of the first order. If x denotes any quantity, the rate of increase 
in x is dx/dt and the rate of decrease in x is — dx/dt ; and consequently 
when the rate of change of x is a function of x, the variables are 
immediately separated and the integration may be performed. The 
constant of integration has to be determined from the initial conditions ; 
the constants inherent in the problem may be given in advance or their 
values may be determined by comparing x and t at some subsequent 
time. The exercises offered below will exemplify the treatment of 
such problems. 

In other physical problems the statement of the question as a differ- 
ential equation is not so direct and is carried out by an examination of 
the problem with a view to stating a relation between the increments 
or differentials of the dependent and independent variables, as in some 
geometric relations already discussed (§ 40), and in the problem of the 
tension in a rope wrapped around a cylindrical post discussed below. 



GEKEBAL IXTRODUCTIOX 



185 



The method may be further illustrated by the derivation of the differ- 
ential equations of the curve of equilibrium of a flexible string or 
chain. Let p be the density of the chain so that pAs is the mass of 
the length As ; let X and Y be the components 
of the force (estimated per unit mass) acting on 
the elements of the chain. Let T denote the 
tension in the chain, and r the inclination of 
the element of chain. From the figure it then 
appears that the components of all the forces 
acting on As are 

(T + AT) cos (r + At) - T cos r + Xp/As = 0, 
(T + AT) sin (t + At) — T sin r + Ypte = ; 

for these must be zero if the element is to be in a position of equi- 
librium. The equations may be written in the form 

A ( T cos r) + Xp As = 0, A ( T sin t) + YpAs = ; 

and if they now be divided by As and if As be allowed to approach 
zero, the result is the two equations of equilibrium 



Y 


P YAs 






T J^V 


pXls 







X 



dx 
ds 



+ P X = 0, 



r tW 



o. 



(4) 



where cos r and sin t are replaced by their values dx/ds and dy/ds. 



If the string is acted on only by forces parallel to a given direction, let the 
y-axis be taken as parallel to that direction. Then the component X will be zero 
and the first equation may be integrated. The result is 



T dx 
ds 



0. 



ds 



C. 



c 



ds 
dx 



This value of T may be substituted in the second equation. There is thus obtained 
a differential equation of the second order 



ds \ dx) 



+ /oF=0 or C 



VI + 



+ pY = 0. 



(40 



T+AT 



If this equation can be integrated, the form of the curve 
of equilibrium may be found. 

Another problem of a different nature in strings is to 
consider the variation of the tension in a rope wound around 
a cylinder without overlapping. The forces acting on the 
element As of the rope are the tensions T and T+AT, the 
normal pressure or reaction E of the cylinder, and the force 
of friction which is proportional to the pressure. It will 
be assumed that the normal reaction lies in the angle A0 and that the coefficient 
of friction is /x so that the force of friction is fiB. The components along the radius 
and alone: the tangent are 




186 DIFFERENTIAL EQUATIONS 

( T + A T) sin A0 - R cos (0A0) - fiR sin (0A<f>) = 0, < < 1, 
(T + AT) cos A0 + R sin (#A0) - /xR cos (#A0) - T = 0. 

Now discard all infinitesimals except those of the first order. It must be borne in 
mind that the pressure R is the reaction on the infinitesimal arc As and hence is 
itself infinitesimal. The substitutions are therefore Td(f> for (T + AT) sin A0, R for 
R cos 0A4>, for R sin 0A<p, and T + dT for (T + AT) cos A£. The equations there- 
fore reduce to two simple equations 

Td<p - R = 0, dT- nR = 0, 

from which the unknown R may be eliminated with the result 

dT=nTdcj> or T = Ce^ or r = T eH>, 

where T is the tension when <p is 0. The tension therefore runs up exponentially 
and affords ample explanation of why a man, by winding a rope about a post, can 
readily hold a ship or other object exerting a great force at the other end of the 
rope. If }jl is 1/3, three turns about the post will hold a force 535 T , or over 25 
tons, if the man exerts a force of a hundredweight. 

84. If a constant mass m is moving along a line under the influence 
of a force F acting along the line, Newton's Second Law of Motion (p. 13) 
states the problem of the motion as the differential equation 

mf=F or ™^ 2 ==F (5) 

of the second order ; and it therefore appears that the complete solution 
of a problem in rectilinear motion requires the integration of this equa- 
tion. The acceleration may be written as 

_ dv _ dv dx _ dv 
J ~~ dt~ dx dt ~ V dx' 

and hence the equation of motion takes either of the forms 

dv „ dv ^ W|x 

m — = F or mv — = F. (o ) 

dt dx 

It now appears that there are several cases in which the first integration 
may be performed. For if the force is a function of the velocity or of 
the time or a product of two such functions, the variables are separated 
in the first form of the equation ; whereas if the force is a function of 
the velocity or of the coordinate x or a product of two such functions, 
the variables are separated in the second form of the equation. 

When the first integration is performed according to either of these 
methods, there will arise an equation between the velocity and either 
the time t or the coordinate x. In this equation will be contained a 
constant of integration which may be determined by the initial condi- 
tions, that is, by the knowledge of the velocity at the start, whether in 



GENERAL INTRODUCTION 187 

time or in position. Finally it will be possible (at least theoretically) 
to solve the equation and express the velocity as a function of the time 
t or of the position x, as the case may be, and integrate a second time. 
The carrying through in practice of this sketch of the work will be 
exemplified in the following two examples. 

Suppose a particle of mass m is projected vertically upward with the velocity V. 
Solve the problem of the motion under the assumption that the resistance of the 
air varies as the velocity of the particle. Let the distance be measured vertically 
upward. The forces acting on the particle are two, — the force of gravity which is 
the weight W = mg, and the resistance of the air which is kv. Both these forces 
are negative because they are directed toward diminishing values of x. Hence 

mf = — mg — kv or m — = — mg — kv, 
dt 

where the first form of the equation of motion has been chosen, although in this 
case the second form would be equally available. Then integrate. 

dV =-dt and \og(g+-v) = --t+C. 
k \ m J m 

9 + -» 

m 

As by the initial conditions v = V when f = 0, the constant C is found from 

k 
I Jr \ k 9 + -v _* t 

log ( g + - V) = - - + C ; hence — = e m 

\ m J m k 

9 + - V 
m 

is the relation between u and t found by substituting the value of C. The solution 

for v gives 

dx tm TT \ -~t m 

V= M=k g+T ) e "'-J 9 ' 

tt m/m \ -~t m 

Hence x = — — t—g + Vje m —-gt+C. 

11 the particle starts from the origin x = 0, the constant C is found to be 
„ m Im , T A , m (m \ / - - A m . 

Hence the position of the particle is expressed in terms of the time and the prob- 
lem is solved. If it be desired to find the time which elapses before the particle 
comes to rest and starts to drop back, it is merely necessary to substitute v = in 
the relation connecting the velocity and the. time, and solve for the time t = T ; 
and if this value of t be substituted in the expression for x, the total distance X 
covered in the ascent will be found. The results are 



i+JLA i,fc)'fi7-ii,(i t M 

mg ) \kj [m \ mg ) \ 



_, m. /„ . k T A -^ /?n\ 2 TA: ^ A k 

T = — loo- 
k 



As a second example consider the motion of a particle vibrating up and down 
at the end of an elastic string held in the field of gravity. By Hooke's Law for 



188 DIFFERENTIAL EQUATIONS 

elastic strings the force exerted by the string is proportional to the extension of 
the string over its natural length, that is, F = kAl. Let I be the length of the string, 
A l the extension of the string just sufficient to hold the weight W = mg at rest so 
that k\l = mg, and let x measured downward be the additional extension of the 
string at any instant of the motion. The force of gravity mg is positive and the 
force of elasticity — k(A l + x) is negative. The second form of the equation of 
motion is to be chosen. Hence 

mv — = mg — k (A J + x) or mv — == — kx, since mg = kAJ. 
dx dx 

Then mvdv = — kxdx or mv 2 = — kx 2 + C. 

Suppose that x = a is the amplitude of the motion, so that when x = a the velocity 
v = and the particle stops and starts back. Then C = ka 2 . Hence 



dx k /—z -„ dx k 

v= — = -y — va 2 — x 2 or — ='\ — 

dt Y m -y/ a 2 _ x 2 Y m 

and sin- 1 - — \ — t + G or x = a sin ( \j— t + C 

a \m \\m 

Now let the time be measured from the instant when the particle passes through 
the position x = 0. Then C satisfies the equation = a sin C and ma y be taken as 
zero. The motion is therefore giv en b y the equation x = asin-Vk/mt and is 
periodic. While t change s by 2 tt Vm/k the particle completes an entire oscilla- 
tion. The time T — 2 tt Vm/k is called the periodic time. The motion considered 
in this example is characterized by the fact that the total force — kx is propor- 
tional to the displacement from a certain origin and is directed toward the origin. 
Motion of this sort is called simple harmonic motion (briefly S. H. M.) and is of 
great importance in mechanics and physics. 

EXERCISES 

1. The sum of S100 is put at interest at 4 per cent per annum under the condition 
that the interest shall be compounded at each instant. Show that the sum will 
amount to $200 in 17 yr. 4 mo., and to $1000 in 56 yr. 

2. Given that the rate of decomposition of an amount x of a given substance is 
proportional to the amount of the substance remaining undecomposed. Solve the 
problem of the decomposition and determine the constant of integration and the 
physical constant of proportionality if x = 5.11 when t = and x = 1.48 when 
t = 40 min. Ans. k = .0309. 

3. A substance is undergoing transformation into another at a rate which is 
assumed to be proportional to the amount of the substance still remaining untrans- 
formed. If that amount is 35.6 when t = 1 hr. and 13.8 when t = 4 hr., determine 
the amount at the start when t = and the constant of proportionality and find 
how many hours will elapse before only one-thousandth of the original amount 
will remain. 

4. If the activity A of a radioactive deposit is proportional to its rate of 
diminution and is found to decrease to \ its initial value in 4 days, show that A 

satisfies the equation A/A = e-°- 173t . 



GENERAL INTRODUCTION 189 

5. Suppose that amounts a and b respectively of two substances are involved in 
a reaction in which the velocity of transformation dx/dt is proportional to the prod- 
uct (a — x) (b — x) of the amounts remaining untransf ormed. Integrate on the 
supposition that a ^ b. 



t 
log b ( a - x ) = , a _ b \ kt . and if ~%% 
a(b-x) 1265 



ft — x b — x 
0.4866 0.2342 
0.3879 0.1354 



determine the product k (a — &) . 

6. Integrate the equation of Ex. 5 if ft = 6, and determine a and k if x = 9.87 
when t = 15 and x = 13.69 when t = 55. 

7. If the velocity of a chemical reaction in which three substances are involved 
is proportional to the continued product of the amounts of the substances remaining, 
show that the equation between x and the time is 



\ft — xj \b — x) \c — x] 



= — kt. where < M 
(a-b)(b-c)(c- ft) \t=0. 

8. Solve Ex. 7 if a = 6 ^ c ; also when a = b = c. Note the very different 
forms of the solution in the three cases. 

9. The rate at which water runs out of a tank through a small pipe issuing 
horizontally near the bottom of the tank is proportional to the square root of the 
height of the surface of the water above the pipe. If the tank is cylindrical and 
half empties in 30 min., show that it will completely empty in about 100 min. 

10. Discuss Ex. 9 in case the tank were a right cone or frustum of a cone. 

11. Consider a vertical column of air and assume that the pressure at any level 
is due to the weight of the air above. Show that p = p e- kh gives the pressure at 
any height h, if Boyle's Law that the density of a gas varies as the pressure be used. 

12. Work Ex. 11 under the assumption that the adiabatic law pocp 1 - 4 repre- 
sents the conditions in the atmosphere. Show that in this case the pressure would 
become zero at a finite height. (If the proper numerical data are inserted, the 
height turns out to be about 20 miles. The adiabatic law seems to correspond 
better to the facts than Boyle's Law.) 

13. Let I be the natural length of an elastic string, let Al be the extension, and 
assume Hooke's Law that the force is proportional to the extension in the form 
Al = klF. Let the string be held in a vertical position so as to elongate under its 
own weight W. Show that the elongation is \TcWl. 

14. The density of water under a pressure of p atmospheres is p = 1 + 0.00004p. 
Show that the surface of an ocean six miles deep is about 600 ft. below the position 
it would have if water were incompressible. . 

15. Show that the equations of the curve of equilibrium of a string or chain are 

ds \ dsj r ' ds\ ds) H 

in polar coordinates, where R and $ are the components of the force along the 
radius vector and perpendicular to it. 



190 DIFFEKENTIAL EQUATIONS 

16. Show that dT + pSds = and T -f- pBN = are the equations of equilib- 
rium of a string if R is the radius of curvature and S and JV are the tangential and 
normal components of the forces. 

17.* Show that when a uniform chain is supported at two points and hangs down 
between the points under its own weight, the curve of equilibrium is the catenary. 

18. Suppose the mass dm of the element ds of a chain is proportional to the pro- 
jection dx of ds on the x-axis, and that the chain hangs in the field of gravity. 
Show that the curve is a parabola. (This is essentially the problem of the shape 
of the cables in a suspension bridge when the roadbed is of uniform linear density ; 
for the weight of the cables is negligible compared to that of the roadbed.) 

19. It is desired to string upon a cord a great many uniform heavy rods of 
varying lengths so that when the chord is hung up with the rods dangling from it 
the rods will be equally spaced along the horizontal and have their lower ends on 
the same level. Required the shape the chord will take. (It should be noted that 
the shape must be known before the rods can be cut in the proper lengths to hang 
as desired.) The weight of the chord may be neglected. 

20. A masonry arch carries a horizontal roadbed. On the assumption that the 
material between the arch and the roadbed is of uniform density and that each 
element of the arch supports the weight of the material above it, find the shape of 
the arch. 

2 1 . In equations (4') the integration may be carried through in terms of quadra- 
tures if pT is a function of y alone ; and similarly in Ex. 15 the integration may be 
carried through if $ = and pE is a function of r alone so that the field is central. 
Show that the results of thus carrying through the integration are the formulas 

Gdy r Cdr/r 



x + C 



J V(/prd V )«-c J ■ 



V(/pB*) 2 -C* 

22. A particle falls from rest through the air, which is assumed to offer a resist- 
ance proportional to the velocity. Solve the problem with the initial conditions 
v — 0, x = 0, t = 0. Show that as the particle falls, the velocity does not increase 
indefinitely, but approaches a definite limit V = mg/k. 

23. Solve Ex. 22 with the initial conditions v — v , x = 0, t = 0, where v is 
greater than the limiting velocity V. Show that the particle slows down as it falls. 

24. A particle rises through the air, which is assumed to resist proportionally to 
the square of the velocity. Solve the motion. Hyperbolic functions are useful. 

25. Solve the problem analogous to Ex. 24 for a falling particle. Show that 
there is a limiting velocity V = vmg/k. If the particle were projected down with 
an initial velocity greater than T, it would slow down as in Ex. 23. 

26. A particle falls towards a point which attracts it inversely as the square of the 
distance and directly as its mass. Find the relation between x and t and determine 
the total time T taken to reach the center. Initial conditions v = 0, x == a, t = 0. 



>l 



2 k , a _ i 2 x — a / -z ^ ■ 7 - 1 (a\ 2 

t = - cos f- Vax — x 2 , T — irk - 



2 a \2 



* Exercises 17-20 should be worked ab initio by the method by which (4) were derived, 
not by applying (4) directly. 



GENERAL INTRODUCTION 191 

27. A particle starts from the origin with a velocity V and moves in a medium 
which resists proportionally to the velocity. Find the relations between velocity 
and distance, velocity and time, and distance and time ; also the limiting distance 
traversed. 

-** -*i 

v = V — fcc/m, v = Ye m , kx = mV(l — e m ), mV/k. 

28. Solve Ex. 27 under the assumption that the resistance varies as Vv. 

29. A particle falls toward a point which attracts inversely as the cube of the 
distance and directly as the mass. The initial conditions are x = a, v = 0, t = 0. 
Show that x 2 = a 2 — kt 2 /a 2 and the total time of descent is T = a 2 /Vk. 

30. A cylindrical spar buoy stands vertically in the water. The buoy is pressed 
down a little and released. Show that, if the resistance of the water and air be 
neglected, the motion is simple harmonic. Integrate and determine the constants 
from the initial conditions x = 0, v = T, t = 0, where x measures the displacement 
from the position of equilibrium. 

31. A particle slides down a rough inclined plane. Determine the motion. Note 
that of the force of gravity only the component mg sin i acts down the plane, 
whereas the component mg cos i acts perpendicularly to the plane and develops the 
force ixmg cos i of friction. Here i is the inclination of the plane and n is the 
coefficient of friction. 

32. A bead is free to move upon a frictionless wire in the form of an inverted 
cycloid (vertex down). Show that the component of the weight along the tangent 
to the cycloid is proportional to the distance of the particle from the vertex. Hence 
determine the motion as simple harmonic and fix the constants of integration by 
the initial conditions that the particle starts from rest at the top of the cycloid. 

33. Two equal weights are hanging at the end of an elastic string. One drops 
off. Determine completely the motion of the particle remaining. 

34. One end of an elastic spring (such as is used in a spring balance) is attached 
rigidly to a point on a horizontal table. To the other end a particle is attached. 
If the particle be held at such a point that the spring is elongated by the amount 
a and then released, determine the motion on the assumption that the coefficient 
of friction between the particle and the table is /j. ; and discuss the possibility of 
different cases according as the force of friction is small or large relative to the 
force exerted by the spring. 

85. Lineal element and differential equation. The idea of a curve 
as made up of the points upon it is familiar. Points, however, have no 
extension and therefore must be regarded not as pieces of a curve but 
merely as positions on it. Strictly speaking, the pieces of a curve are 
the elements As of arc ; but for many purposes it is convenient to re- 
place the complicated element As by a piece of the tangent to the curve 
at some point of the arc As, and from this point of view a curve is made 
up of an infinite number of infinitesimal elements tangent to it. This 
is analogous to the point of view by which a curve is regarded as made 




192 DIFFERENTIAL EQUATIONS 

up of an infinite number of infinitesimal chords and is intimately related 
to the conception of the curve as the envelope of its tangents (§ 65). 
A point on a curve taken with an infinitesimal portion of the tangent 
to the curve at that point is called a lineal element of the curve. These 
concepts and definitions are clearly equally available in two or three 
dimensions. For the present the curves under dis- 
cussion will be plane curves and the lineal elements 
will therefore all lie in a plane. • /(x,yp) 

To specify any particular lineal element three 
coordinates x, y, p will be used, of which the two (x, y) determine the 
point through which the element passes and of which the third p is 
the slope of the element. If a curve f(x, y) = is given, the slope at 
any point may be found by differentiation, 

p = d JL=-f/f, (6) 

ax ex' cy v 7 

and hence the third coordinate p of the lineal elements of this particular 
curve is expressed in terms of the other two. If in place of one curve 
f(x, y) = the whole family of curves f(x, y) = C, where C is an 
arbitrary constant, had been given, the slope p would still be found 
from (6), and it therefore appears that the third coordinate of the lineal 
elements of such a family of curves is expressible in terms of x and y. 
In the more general case where the family of curves is given in the 
unsolved form F(x, y, C) = 0, the slope p is found by the same formula 
but it now depends apparently on C in addition to on x and y. If, how- 
ever, the constant C be eliminated from the two equations 

F{x,y,C) = and °£ + c -£p = 0, (7) 

there will arise an equation $ (x, y, p) — which connects the slope p 
of any curve of the family with the coordinates (x, y) of any point 
through which a curve of the family passes and at which the slope of 
that curve is p. Hence it appears that the three coordinates (x, y,j)) of 
the lineal elements of all the curves of a family are connected by an equa- 
tion 3> (x, y, p) = 0, just as the coordinates (x, y, z) of the points of a 
surface are connected by an equation <£ (x, y, z) = 0. As the equation 
<£(x, y, z) = is called the equation of the surface, so the equation 
$ (x, y, p) = is called the equation of the family of curves ; it is, how- 
ever, not the finite equation F(x, y, C) = but the differential equation 
of the family, because it involves the derivative p = dy/clx of y by x 
instead of the parameter C. 



GENERAL INTRODUCTION 193 

As an example of the elimination of a constant, consider the case of the parabolas 

y 2 = Cx or y 2 /x = C. 

The differentiation of the equation in the second form gives at once 

— y 2 /x 2 + 2 yp/x = or y = 2 xp 

as the differential equation of the family. In the unsolved form the work is 

2 yp = C, y 2 = 2 ypx, y = 2xp. 

The result is, of course, the same in either case. For the family here treated it 
makes little difference which method is followed. As a general rule it is perhaps 
best to solve for the constant if the solution is simple and leads to a simple form 
of the function /(&, y) ; whereas if the solution is not simple or the form of the 
function is complicated, it is best to differentiate first because the differentiated 
equation may be simpler to solve for the constant than the original equation, or 
because the elimination of the constant between the two equations can be con- 
ducted advantageously. 

If an equation <£ (x, y,p) = connecting the three coordinates of the 
lineal element be given, the elements which satisfy the equation may 
be plotted much as a surface is plotted ; that is, a pair of values (x, y) 
may be assumed and substituted in the equation, the equation may then 
be solved for one or more values of p, and lineal elements with these 
values of p may be drawn through the point (x, ?/). In this manner the 
elements through as many points as desired may be found. The de- 
tached elements are of interest and significance chiefly from the fact 
that they can be assembled into curves, — in fact, into the curves of a 
family F (x, y, C) = of which the equation <£ (x, y, p) = is the differ- 
ential equation. This is the converse of the problem treated above and 
requires the integration of the differential equation <£ (x, y, p) = for its 
solution. In some simple cases the assembling may be accomplished 
intuitively from the geometric properties implied in the equation, in 
other cases it follows from the integration of the equation by analytic 
means, in other cases it can be done only approximately and by methods 
of computation. 

As an example of intuitively assembling the lineal elements into curves, take 

V)*' — y 2 
$ (x, y, p) = y 2 p 2 + y 2 — r 2 = or p = ± ■ ■ 



The quantity V /'' 2 — y 2 may be interpreted as one leg of a right triangle of which 
y is th e other leg and r the hypotenuse. The slope of the hypotenuse is then 
± l// vV 2 — y 2 according to the position of the figure, and the differential equation 
$ (x, y, p) = states that the coordinate p of the lineal element which satisfies it 
is the negative reciprocal of this slope. Hence the lineal element is perpendicular 
to the hypotenuse. It therefore appears that the lineal elements are tangent to cir- 
cles of radius r described about points of the x-axis. The equation of these circles is 



194 DIFFEBENTIAL EQUATIONS 

(x — C) 2 + y 2 = r 2 , and this is therefore the integral of the differential equation. 
The correctness of this integral may be checked by direct integration. Eor 



dy vr 2 — y 2 ydy , r-z ^ 

p = — = ± — or J = dx or Vr 2 - y 2 = x — C. 

dx y Vr 2 — ij 1 

86. In geometric problems which relate the slope of the tangent of a 
curve to other lines in the figure, it is clear that not the tangent but 
the lineal element is the vital thing. Among such problems that of the 
orthogonal trajectories (or trajectories under any angle) of a given family 
of curves is of especial importance. If two families of curves are so 
related that the angle at which any curve of one of the families cuts 
any curve of the other family is a right angle, then the curves of either 
family are said to be the orthogonal trajectories of the curves of the 
other family. Hence at any point (x, y) at which two curves belonging 
to the different families intersect, there are two lineal elements, one 
belonging to each curve, which are perpendicular. As the slopes of two 
perpendicular lines are the negative reciprocals of each other, it follows 
that if the coordinates of one lineal element are (x, y, p) the coordinates 
of the other are (x, y, — 1/jj) ; and if the coordinates of the lineal ele- 
ment (x, y, p) satisfy the equation <£ (x, y, p) = 0, the coordinates of the 
orthogonal lineal element must satisfy ®(x, y, — 1/p) = 0. Therefore 
the rule for finding the orthogonal trajectories of the curves F(x, y, C)= 
is to find first the differential equation $>(x, y,j)) = of 'the family ', then 
to replace p by — 1/p to find the differential equation of the orthogonal 
family, and finally to integrate this equation to find the family. It may 
be noted that if F(z) = X (x, y) -f- iY(x, y) is a function of z = x -f iy 
(§ 73), the families X(x, y) = C and Y(x, y) = K are orthogonal. 

As a problem in orthogonal trajectories find the trajectories of the semicubical 
parabolas (x — C) 3 = y 2 . The differential equation of this family is found as 

3 (x - C) 2 = 2 yp, x- C = (f yp)i, (| yp)§ = y 2 or \p = yi. 

This is the differential equation of the given family. Replace p by — 1/p and 
integrate : 

= ?/3 or 1 + - pi/3 =0 or dx + - 2/ ¥ dy = 0, and x + - y* = C. 

Sp 2 2 8 

Thus the differential equation and finite equation of the orthogonal family are found. 
The curves look something like parabolas with axis horizontal and vertex toward 
the right. 

Given a differential equation 3> (x, y, p) = or, in solved form, 
p = <f> (x, y) ; the lineal element affords a means for obtaining graphically 
and numerically an approximation to the solution which passes through 



GEKEBAL LNTKODUCTION 



195 



an assigned point P (x Q , y ). For the value p Q of p at this point may be 
computed from the equation and a lineal element P Q P X may be drawn, 
the length being taken small. As the lineal element is tangent to the 
curve, its end point will not lie upon the curve but will depart from it 
by an infinitesimal of higher order. Next the slope p 1 of the lineal 
element which satisfies the equation and passes 
through P may be found and the element PJP^ 
may be drawn. This element will not be tangent 
to the desired solution but to a solution lying near 
that one. Xext the element P. 2 P S may be drawn, 
and so on. The broken line PJP x PjP z • • • is clearly 

an approximation to the solution and will be a better approximation 
the shorter the elements P { Pi +1 are taken. If the radius of curvature 
of the solution at P Q is not great, the curve will be bending rapidly and 
the elements must be taken fairly short in order to get a fair approx- 
imation ; but if the radius of curvature is great, the elements need not 
be taken so small. (This method of approximate graphical solution 
indicates a method which is of value in proving by the method of 
limits that the equation p = <f> (x, y) actually has a solution ; but that 
matter will not be treated here.) 




Let it be required to plot approximately that solution of yp + x = which 
passes through (0, 1) and thus to find the ordinate for x = 0.5, and the area under 
the curve and the length of the curve to this point. Instead of assuming the lengths 
of the successive lineal elements, let the 
lengths of successive increments 8x of 
x be taken as 8x — 0.1. At the start 
x = 0, y = 1, and from p = — x/ij it 
follows that p = 0. The increment 8y 
of y acquired in moving along the tan- 
gent is 8y = p8x = 0. Hence the new 
point of departure (x x , y x ) is (0.1, 1) and 
the new slope is p x = — x l /y 1 = — 0.1. 
The results of the work, as it is contin- 
ued, may be grouped in the table. Hence it appears that the final ordinate is 
y = 0.90. By adding up the trapezoids the area is computed as 0.48, and by find- 
ing the elements 8s = Vsx' 2 + 8y 2 the length is found as 0.51. Now the particular 
equation here treated can be integrated. 

yp + x = 0, ydy + xdx = 0, x 2 + y' 2 = C, and hence x 2 + y 2 = 1 

is the solution which passes through (0, 1). The ordinate, area, and length found 
from the curve are therefore 0.87, 0.48, 0.52 respectively. The errors in the 
approximate results to two places are therefore respectively 3, 0, 2 per cent. If 8x 
had been chosen as 0.01 and four places had been kept in the computations, the 
errors would have been smaller. 



i 


8x 


sy 


Xi 


Vi 


Vi 









0. 


1.00 


0. 


1 


0.1 


0. 


0.1 


1.00 


-0.1 


2 


0.1 


-0.01 


0.2 


0.99 


-0.2 


3 


0.1 


-0.02 


0.3 


0.97 


- 0.31 


4 


0.1 


- 0.03 


0.4 


0.94 


- 0.43 


5 


0.1 


-0.04 


0.5 


0.90 





196 DIFFERENTIAL EQUATIONS 

EXERCISES 

1. In the following cases eliminate the constant C to find the differential equa- 
tion of the family given : 



(a) x 2 = 2Cy + C 2 , (/3) y = Cx + Vl - C\ 

(7) x 2 — ij 2 = Cx, (5) y = x tan (x + C), 

(e ) -*— + ^ = 1, An* (^Y + ( * 2 -^~ (a2 " 6 ' 2) d2/ 

v ' a 2 - C b 2 - C \dx) xy 



-1 = 0. 
xy dx 



2. Plot the lineal elements and intuitively assemble them into the solution : 

{a) yp .+ x "= 0, (|8) xp - y = 0, (7) r — = 1. 

Check the results by direct integration of the differential equations. 

3. Lines drawn from the points (± c, 0) to the lineal element are equally in- 
clined to it. Show that the differential equation is that of Ex. 1 (e). What are the 
curves ? 

4. The trapezoidal area under the lineal element equals the sectorial area formed 
by joining the origin to the extremities of the element (disregarding infinitesimals 
of higher order), (a) Find the differential equation and integrate. ((8) Solve the 
same problem where the areas are equal in magnitude but opposite in sign. What 
are the curves ? 

5. Find the orthogonal trajectories of the following families. Sketch the curves. 

(a) parabolas y 2 = 2 Cx, Ans. ellipses 2x 2 + y 2 = C. 

(/3) exponentials y = Ce l ' x , Ans. parabolas \~ky 2 + x = C. 

(7) circles (x — C) 2 + y 2 = a' 2 , Ans. tractrices. 

(5 ) X 2 _ y2 = C a ) ( e ) Cy 2 = x 3 , (f ) at + ifi = Ci. 

6. Show from the answer to Ex. 1 (e) that the family is self -orthogonal and 
illustrate with a sketch. From the fact that the lineal element of a parabola makes 
equal angles with the axis and with the line drawn to the focus, derive the differ- 
ential equation of all coaxial confocal parabolas and show that the family is self- 
orthogonal. 

7. If * (x, y, p) = is the differential equation of a family, show 

*L v ,p^\ = o and i L y ,l+^) = o 

\ 1 + rap/ \ 1 — mpj 

are the differential equations of the family whose curves cut those of the given 
family at tan- 1 m. What is the difference between these two cases ? 

8. Show that the differential equations 



$ 



(|,,,) = and .(-^.r I# )-0 



define orthogonal families in polar coordinates, and write the equation of the family 
which cuts the first of these at the constant angle tan- 1 m. 

9. Find the orthogonal trajectories of the following families. Sketch. 

(a) r = eC</>, (p) r = C(l - cos0), (7) r = C<p, (8) r 2 = C 2 cos 2 0. 



GENERAL INTRODUCTION 197 

10. Recompute the approximate solution of yp + x = under the conditions of 
the text but with 8x = 0.05, and carry the work to three decimals. 

11. Plot the approximate solution of p = xy between (1, 1) and the y-axis. Take 
8x = — 0.2. Find the ordinate, area, and length. Check by integration and 
comparison. 

12. Plot the approximate solution of p — — x through (1, 1), taking 8x = 0.1 and 
following the curve to its intersection with the x-axis. Find also the area and the 
length. 



13. Plot the solution of p = Vx' 2 + if- from the point (0, 1) to its intersection 
with the x-axis. Take 8x — — 0.2 and find the area and length. 

14. Plot the solution of p = s which starts from the origin into the first quad- 
rant (s is the length of the arc). Take 5x = 0.1 and carry the work for five steps 
to find the final ordinate, the area, .and the length. Compare with the true integral. 

87. The higher derivatives ; analytic approximations. Although a 
differential equation 3> (x, y, y') = does not determine the relation 
between x and y without the application of some process equivalent to 
integration, it does afford a means of computing the higher derivatives 
simply by differentiation. Thus 

t/<E> c<& t d& . d& ,. . 

-T- = — + — y' + — y" = 
ax ex cy cy' 

is an equation which may be solved for y" as a function of x, y, y' ; 
and y" may therefore be expressed in terms of x and y by means of 
3?(x, y, y') = 0. A further differentiation gives the equation 

ax- ex- excy dxcy' cy 1 c y c y 

dy' 2 J dy J oy { J 

which may be solved for y'" in terms of x, y, y', y"; and hence, by the 
preceding results, y'" is expressible as a function of x and y ; and so 
on to all the higher derivatives. In this way any property of the inte- 
grals of ®(x, y, y') = which, like the radius of curvature, is expressi- 
ble in terms of the derivatives, may be found as a function of x and y. 
As the differential equation $(x, y, y') = defines y' and all the 
higher derivatives as functions of x, y, it is clear that the values of the 
derivatives may be found as y , y' ', g/J", • • • at any given point (x Q , y ). 
Hence it is possible to write the series 

V = y + !/o ( x ~ x o) + i Vo ( x - x of + * v7 ( x ~ x of + • • '• ( 8 ) 
If this power series in x — x converges, it defines y as a function of 
x for values of x near x Q ; it is indeed the Taylor development of the 



198 DIFFERENTIAL EQUATIONS 

function y (§ 167). The convergence is assumed. Then 

y = ?/o + Vo(x - O + Wi r (? - x of +"'• 
It may be shown that the function y defined by the series actually 
satisfies the differential equation &(x, y, y') = 0, that is, that 

Q(x)=<P[x, y + y' (x-x^ + \y'i (x-x Q ) 2 + - ■ . } y' Q + y' Q ' ( x -x Q ) + - ■ -] = 

for all values of x near x Q . To prove this accurately, however, is beyond 
the scope of the present discussion ; the fact may be taken for granted. 
Hence an analytic expansion for the integral of a differential equa- 
tion has been found. 

As an. example of computation with higher derivatives let it be required to deter- 
mine the radius of curvature of that solution of y' — tan (y/x) which passes through 
(1, 1). Here the slope y' {X 1} at (1, 1) is tan 1 == 1.557. The second derivative is 

r = *¥- = ?-\»*y- = «**-**-*. 

dx dx x x x 2 

From these data the radius of curvature is found to be 

B= (l + y /2 ) ? = sec y_J^_ > R seci 1 = 3.250. 

y" x xy' — y tan 1 — 1 

The equation of the circle of curvature may also be found. For as y"^ 1} is positive, 
the curve is concave up. Hence (1 — 3.250 sin 1, 1 + 3.250 cos 1) is the center of 
curvature ; and the circle is 

(x + 1. 735) 2 + (y - 2.757) 2 = (3.250) 2 . 

As a second example let four terms of the expansion of that integral of 
a; tan?/' = y which passes through (2, 1) be found. The differential equation may- 
be solved ; then 

dy = tan-i/^ d2y - Xy/ ~~ V 



dx \x) dx 2 x' 1 + y 2 

d s y __ (x 2 + y 2 ) (x — 1) y" + (3 y 2 — x 2 ) y' — 2 xyy' 2 + 2xy 
dx* ~ (x 2 + y 2 ) 2 

Now it must be noted that the problem is not wholly determinate ; for y' is multi- 
ple valued and any one of the values for tan -1 \ may be taken as the slope of a 
solution through (2, 1). Suppose that the angle be taken in the first quadrant ; then 
tan- 1 \ = 0.462. Substituting this in y", we find y^ 1) =— 0.0152 ; and hence may 
be found y"^ X) = 0.110. The series for y to four terms is therefore 

y = 1 + 0.462 (x - 2) - 0.0076 (x - 2) 2 + 0.018 (x - 3) 3 . 

It may be noted that it is generally simpler not to express the higher derivatives in 
terms of x and y, but to compute each one successively from the preceding ones. 

88. Picard has given a method for the integration of the equation 
y' — <fj(x, y) by successive approximations which, although of the highest 
theoretic value and importance, is not particularly suitable to analytic 



GENERAL INTRODUCTION 199 

uses in finding an approximate solution. The method is this. Let the 
equation y' = <f> (x, y) be given in solved form, and suppose (x , y ) is 
the point through which the solution is to pass. To find the first 
approximation let y be held constant and equal to y , and integrate the 
equation y' = <f>(x, ?/ ). Thus 

dy = <f> {x, y Q ) dx; y = y + / t£ (x, y ) dx = f x (x), (9) 

where it will be noticed that the constant of integration has been chosen 
so that the curve passes through (x Q , y Q ). For the second approximation 
let y have the value just found, substitute this in <f>(x, y), and integrate 
again. Then 



Jx Q L «^ 



y<>+ \ <t>\x,y< i + \ <t>( x ,y*) dx 



dx=f 2 (x). (9') 



^Yith this new value for y continue as before. The successive deter- 
minations of y as a function of x actually converge toward a limiting 
function which is a solution of the equation and which passes through 
(x Q , y Q ). It may be noted that at each step of the work an integration 
is required. The difficulty of actually performing this integration in 
formal practice limits the usefulness of the method in such cases. It is 
clear, however, that with an integrating machine such as the integraph 
the method could be applied as rapidly as the curves <f> (x, f { (x)) could 
be plotted. 

To see how the method works, consider the integration of y' = x + y to find the 
integral through (1, 1). For the first approximation y = 1. Then 

dy = {x + 1) dx, y = ±x 2 + x+C, y = ix 2 + x- \ =f 1 {x). 

From this value of y the next approximation may be found, and then still another t 

dy = [x + (i^ + x - i)]dx, y = % x s + x 2 - \x + i =f,(x), 
dy = [x +/ 2 (x)] dx, y = ^x± + § x s + \x 2 + Ix + ^. 

In this case there are no difficulties which would prevent any number of appli- 
cations of the method. In fact it is evident that if y' is a polynomial in x and y, the 
result of any number of applications of the method will be a polynomial in x. 

The method of undetermined coefficients may often be employed to 
advantage to develop the solution of a differential equation into a 
series. The result is of course identical with that obtained by the 
application of successive differentiation and Taylor's series as above ; 
the work is sometimes shorter. Let the equation be in the form 
y' = <f> (x, ?/) and assume an integral in the form 

y = y<> + a i ( x - x o) + «2 ( x - x o) 2 + a s ( x - x oT + ■■■> ( 10 ) 



200 DIFFERENTIAL EQUATIONS 

Then <£ (x, y) may also be expanded into a series, say, 

<£(a>, y) = A + A x (x - x ) + ^ 2 (x - x f + ^ (a; - x Q f -f • • - 

But by differentiating the assumed form for y we have 

2/' = «! + 2 a a (aj - a; ) + 3 a 8 (a; - a? ) 2 + 4 a 4 (a; - x^f + • • •• 

Thus there arise two different expressions as series in x — x Q for the 
function y', and therefore the corresponding coefficients must be equal. 
The resulting set of equations 

a 1 = A Q , 2a 2 = A v Sa^ = A 2 , ±a^ = A B , • • , (11) 

may be solved successively for the undetermined coefficients a v a 2 , a g , 
a Aj • • • which enter into the assumed expansion. This method is partic- 
ularly useful when the form of the differential equation is such that 
some of the terms may be omitted from the assumed expansion (see 
Ex. 14). 

As an example in the use of undetermined coefficients consider that solution of 
the equation y' — Vx 2 + Sy 2 which passes through (1, 1). The expansion will pro- 
ceed according to powers of x — 1, and for convenience the variable may be changed 
to t = x — 1 so that 

dy 



V(t + l) 2 + Sy\ y = 1 + %*-+ a/ 2 + a z t* + a 4 t± + • • ■ 

are the equation and the assumed expansion. One expression for y' is 

y' = a x + 2 a 2 t + 3 a 3 t 2 + 4 a±t s + • • • . 
To find the other it is necessary to expand into a series in t the expression 



f = V(l + t) 2 + 3 (1 + aj + a 2 t 2 + a^f 



If this had to be done by Maclaurin's series, nothing would be gained over the 
method of § 87 ; but in this and many other cases algebraic methods and known 
expansions may be applied (§ 32) . First square y and retain only terms up to the 
third power. Hence 



2/ / = 2Vl + i(l + 3a 1 )i + i(l + 6a 2 + 3 a 2 ) t 2 + f {a x a 2 + a 3 ) t s . 
Now let the quantity under the radical be called 1 + h and expand so that 

f = 2 Vl + h = 2 (1 + \ h - i h 2 + xV /i 3 ). 
Finally raise h to the indicated powers and collect in powers of t. Then 
t 



?/' = 2+i(l + 3a 1 ) 



+ i(l + 6a 2 + 3a 2 ) 



* 3 

+ I K« 2 + a s) 

-^(1 + 3o t )(l +603,4- Sof) 

+ e 1 T(l + 3a 1 ) 3 



GENERAL INTRODUCTION 201 

Hence the successive equations for determining the coefficients are a x = 2 and 

2 a 2 = I (1 + 3 a x ) or a. 2 = |, 

Sa 3 = i(l + 6« 2 + 3af) - T V(1 + 3^ or a 3 = if, 

4 a 4 = | (a^ + a 8 ) - T \ (1 + 3 a x ) (1 + 6 a 2 + 3 af) + £ (1 + 3 atf or a 4 = ft|. 

Therefore to five terms the expansion desired is 

y = i + 2 (x - 1) + | (x - 1)2 + if (x - 1)« + m (x - 1)K 

The methods of developing a solution by Taylor's series or by un- 
determined coefficients apply equally well to equations of higher order. 
For example consider an equation of the second order in solved form 
y" = <£ (x, y, y') and its derivatives 

J ex cy J cy' J 

y * = ^ + 2^y< + 2^y" + ^y» + 2?$-,yy 
J ex 2 excy J cxcij J cy 2 J cycy' J J 

dif * cy J cy' * 

Evidently the higher derivatives of y may be obtained in terms of x, 
y. y' ; and y itself may be written in the expanded form 

y = y + y'o( x - x o) + h y'o( x - x T + 4 yo-( x - x o) s / 12 ) 

~T" 24 yQ \ X X o) I ' ' ' J 

where any desired values may be attributed to the ordinate y at which 
the curve cuts the line x = x , and to the slope y of the curve at that 
point. Moreover the coefficients y' ', y ", • ■ • are determined in such a way 
that they depend on the assumed values of y and y . It therefore is 
seen that the solution (12) of the differential equation of the second 
order really involves two arbitrary constants, and the justification of 
writing it as F(x, y, C , C 2 ) = is clear. 

In following out the method of undetermined coefficients a solution 
of the equation would be assumed in the form 

y = y + y'o( x - *„) + a -i ( x - x d' 2 + % ( x - x o) s + r \ ( x - x oY + • • ■ > ( 13 ) 

from which y' and y" would be obtained by differentiation. Then if the 
series for y and y' be substituted in y" = cf>(x, y, y') and the result 
arranged as a series, a second expression for y" is obtained and the 
comparison of the coefficients in the two series will afford a set of equa- 
tions from which the successive coefficients may be found in terms of 
y and y by solution. These results may clearly be generalized to the 
case of differential equations of the nth. order, whereof the solutions 
will depend on n arbitrary constants, namely, the values assumed for 
y and its first n — 1 derivatives when x = x n . 



202 DIFFERENTIAL EQUATIONS 

EXERCISES 

1. Find the radii and circles of curvature of the solutions of the following equa- 
tions at the points indicated : 

(a) y' = Vx 2 + V 2 at (0, 1), (p) yy' + x = at (x , y ). 

2. Find ij'H X) = (5 V2 - 2)/4 if ^ = Vx 2 + y 2 . 

3. Given the equation ?/ 2 2/' 3 + x?/?/' 2 — 2/?/ + x 2 = of the third degree in y' so 
that there will be three solutions with different slopes through any ordinary point 
(x, y). Find the radii of curvature of the three solutions through (0, 1). 

4. Find three terms in the expansion of the solution of y' = ew about (2, \). 

5 . Find four terms in the expansion of the solution ofy= log sin xy about Q?r, 1). 

6. Expand the solution of y' = xy about (1, y Q ) to five terms. 

7. Expand the solution of y / = tan (y/x) about (1, 0) to four terms. Note that 
here x should be expanded in terms of y, not y in terms of x. 

8. Expand two of the solutions of y 2 y' 3 + xyy" 1 — yy' + x 2 = about (— 2, 1) 
to four terms. 

9. Obtain four successive approximations to the integral of y'=xy through (1, 1). 

10. Find four successive approximations to the integral of y' = x + y through 
(0, 2/ ). 

1 1 . Show by successive approximations that the integral of y / = y through (0, y ) 
is the well-known y = y e x . 

12. Carry the approximations to the solution of y' = — x/y through (0, 1) as 
far as you can integrate, and plot each approximation on the same figure with the 
exact integral. 

13. Find by the method of undetermined coefficients the number of terms indi- 
cated in the expansions of the solutions of these differential equations about the 
points given : 

(a) y' = Vx + y, five terms, (0, 1), (/3) y' = Vx + y, four terms, (1, 3), 
(7) V' ■= x + y, n terms, (0, y ), (8) if = Vx 2 + y 1 , four terms, (f, i). 

14. If the solution of an equation is to be expanded about (0, y ) and if the 

change of x into — x and y' into — y' does not alter the equation, the solution is 

necessarily symmetric with respect to the ?/-axis and the expansion may be assumed 

to contain only even powers of x. If the solution is to be expanded about (0, 0) 

and a change of x into — x and y into — y does not alter the equation, the solution 

is symmetric with respect to the origin and the expansion may be assumed in odd 

powers. Obtain the expansions to four terms in the following cases and compare 

the labor involved in the method of undetermined coefficients with that which 

would be involved in performing the requisite six or seven differentiations for the 

application of Maclaurin's series : 

x 
(a) y' = — ' about (0, 2), (/3) y' — sin xy about (0, 1), 

Vx 2 + y' 2 
(7) y' = e x v about (0, 0), (5) y' = x s y + xy 3 about (0, 0). 

15. Expand to and including the term x 4 : 

(a) y" = y' 2 + xy about x = 0, y = a , y' = a x (by both methods), 

0, y = a , y' =-a (by unci, coeffs.). 



CHAPTER VIII 

THE COMMONER ORDINARY DIFFERENTIAL EQUATIONS 

89. Integration by separating the variables. If a differential equa- 
tion of the first order may be solved for y 1 so that 

y' = *(*,y) or M(x, y)dx + N(x,y)dy = (1) 

(where the functions cf>, M, X are single valued or where only one spe- 
cific branch of each function is selected in case the solution leads to 
multiple valued functions), the differential equation involves only the 
first power of the derivative and is said to be of the first degree. If, 
furthermore, it so happens that the functions <f>, M, N are products of 
functions of x and functions of y so that the equation (1) takes the form 

y f = ^{x)^ 2 (y) or M^x) M&j) dx + N^x) NJ#) dy = 0, (2) 

it is clear that the variables may be separated in the manner 

-^- = d>MJx or ¥i£Ldx + m&d v = (2') 

and the integration is then immediately performed by integrating each 
side of the equation. It was in this way that the numerous problems 
considered in Chap. VII were solved. 

As an example consider the equation yy' + xy' 2 = x. Here 

ydy + x (y 2 — 1) dx = or V V + xdx - 0, 
y 2 - 1 

and \ log (y 2 -l) + ix 2 =C or {y 2 — 1) e^ 2 = C. 

The second form of the solution is found by taking the exponential of both sides 
of the first form after multiplying by 2. 

In some differential equations (1) in which the variables are not 
immediately separable as above, the introduction of some change of 
variable, whether of the dependent or independent variable or both, 
may lead to a differential equation in which the new variables are sepa- 
rated and the integration may be accomplished. The selection of the 
proper change of variable is in general a matter for the exercise of 
ingenuity ; succeeding paragraphs, however, will point out some special 

203 



204 * DIFFERENTIAL EQUATIONS 

types of equations for which a definite type of substitution is known 
to accomplish the separation. 



As an example consider the equation xdy — ydx = x Vx 2 + y 2 dx, where th e varia- 
bles are clearly not separable without substitution. The presence of Vx 2 + y 2 
suggests a change to polar coordinates. The work of finding the solution is : 

x = r cos 6, y = r sin 0, dx = cos Odr — r sin 6d6, dy = sin Odr + r cos 6d0 ; 



then xdy — ydx = r 2 d6, x Vx 2 + y 2 dx = r 2 cos Od (r cos 0) . 

Hence the differential equation may be written in the form 

r 2 d0 = r 2 cos Od (r cos 0) or sec OdO = d(r cos 0), 

1 I Hri 

and log tan (10 + \ie\ — r cos + C or log — = x + C. 

cos 



y X 2 + V 2 + V 

Hence = Ce x (on substitution for 0) . 

x 

Another change of variable which works, is to let y = vx. Then the work is 



x (vdx + xdv) — vxdx = x 2 V 1 + u 2 dx or dv = V 1 + v 2 dx. 

dv 
Then , = dx, sinh- 1 ^ = x + C, 2/ = x sinh (x + C) . 

VI + v 2 

This solution turns out to be shorter and the answer appears in neater form than 
before obtained. The great difference of form that may arise in the answer when 
different methods of integration are employed, is a noteworthy fact, and renders a 
set of answers practically worthless ; two solvers may frequently waste more time 
in trying to get their answers reduced to a common form than each would spend in 
solving the problem in two ways. 

90. If in the equation y' = </> (x, y) the function <f> turns out to be 
<j>Q//x), a function of y/x alone, that is, if the functions M and N are 
homogeneous functions of x, y and of the same order (§ 53), the differ- 
ential equation is said to be homogeneous and the change of variable 
y = vx or x = vy will always result in separating the variables. The 
statement may be tabulated as : 



if ^- = Jt\ 

dx \x 



or x = vy. x 7 



A sort of corollary case is given in Ex. 6 below. 

As an example take y \1 + eydx + e& (y — x)dy — 0, of which the homogeneity 
is perhaps somewhat disguised. Here it is better to choose x = vy. Then 

(1 + e v ) dx + e v (1 — v) dy = and dx = vdy + ydv. 

dv 1 + e v 
Hence (v + e v ) dy + y (1 + e v ) dv = or — -\ dy = 0. 

y v + e v 

X 

Hence log y + log (v + e v ) = C or x + y& = C. 



COMMOKEK OBDINARY EQUATIONS 205 

If the differential equation may be arranged so that 

f x + X l (x)y = X i (x)y* or g + Yfy) x = Yfy) *, (4) 

where the second form differs from the first only through the inter- 
change of x and y and where X 1 and X 2 are functions of x alone and 
Y and Y 2 functions of y, the equation is called a Bernoulli equation; and 
in particular if n = 0, so that the dependent variable does not occur on 
the right-hand side, the equation is called linear. The substitution 
which separates the variables in the respective cases is 

y = ve -f*iWte or x = veS T ^ dv . (5) 

To show that the separation is really accomplished and to find a general 
formula for the solution of any Bernoulli or linear equation, the sub- 
stitution may be carried out formally. For 

*!L.tof-S**>-vXjrJ'*. 

ax ax 1 

The substitution of this value in the equation gives 

C Il e - fx.dx = X n e - nfx t dx or ^ = J/1" »>/**« fa. 

ax Z v n 2 

Hence v 1 ~ n = (1 — n) I X 2 e a ~ n) f Xldx dx, when n =£= 1* 

or if- n =(l- n) e (n -D A^| C X/ 1 - n) ! x ^ lx dx\ . (6) 

There is an analogous form for the second form of the equation. 

The equation (x 2 y z + xy) dy = dx may be treated by this method by writing it as 

dx 

yx = y s x 2 so that F, = — y. F„ = y 3 . n = 2. 

dy 

Then let x = ve'f' ydy = ve* v \ 

rp, dX dV ij/2 I j,2 lj/2 dV |y2 

Then ■ yx = — e~ + vye — yve z = — e" 

dy dy . dy 

and — e' 2 = yH 2 e^ or — = y 3 e 2 dy, 

dy v 2 

and - - =(y 2 - 2) e^ + C or i = 2 - ?/ 2 + Ce" ^ . 

v x 

This result could have been obtained by direct substitution in the formula 

X l-n = (1 _ n)e (-l)/^^ 7 /-)/W^ ( 

but actually to carry the method through is far more instructive. 

* If fi=l, the variables are separated in the original equation. 



206 DIFFERENTIAL EQUATIONS 

EXERCISES 

1. Solve the equations (variables immediately separable) : 

(a) (1 + x)y + (1 — y)xy' — 0, Ans. xy = CeF-'Jf. 

(P) a (xdy + 2 ydx) = xydy, (7) Vl - x 2 dy + Vl - y 2 dx = 0, 

(5) (1 + y 2 ) dx - (y + Vl + y){l + x)l dy = 0. 

2. By various ingenious changes of variable, solve : 

(a) (x + y) 2 y' = a 2 , ^Lns. x + y — a tan (y/a + C). 

(/3) (x — y 2 ) dx + 2 xydy — 0, (7) xdy — ydx — (x 2 + y 2 ) dx, 

(5) y' = x-y, (e) yy' + y 2 + x + 1 = 0. 

3 . Solve these homogeneous equations : 

(a) (2 Vxy — x) y'. + y = 0, Ans. Vx/y 4- log y = C. 

I 
(p) xe x + y — xy' = 0, ^4ws. y 2 (x 2 + y 2 ) = Cx 6 . 

(7) (x 2 + 2/ 2 ) dy = xydx, (5) xy' — y = Vx 2 + y 2 . 

4. Solve these Bernoulli or linear equations : 

(or) y' + y/x = y 2 , Ans. xy log Cx + 1 = 0. 

(/3) ?/' — y esc x = cosx — 1, -4ns. y = sin x + C tan |x. 

(7) %y' + y = y 2 log x, Ans. y- 1 = log x + 1 + Cx. 

(8) (1 + ?/ 2 ) dx + (tan- 1 y — x) dy, (e) ydx + (ax 2 y n — 2x)dy = 0, 
(O xy' — ay = x + l, (rj) yy' + \ y 2 = cos x. 

5. Show that the substitution y = vx always separates the variables in the 
homogeneous equation y' = <f> (y/x) and derive the general formula for the integral. 

6. Let a differential equation be reducible to the form 

dy / a x x + b x y + cA G4& 2 — a^ ^ 0, 

dx ~~ \a 2 x + 6 2 y + c 2 / ' or a^ - a 2 6 1 = 0. 

In case a t b 2 — a 2 \ ^ 0, the two lines a x x + ^y + c x = and a 2 x + b 2 y + c 2 = 
will meet in a point. Show that a transformation to this point as origin makes 
the new equation homogeneous and hence soluble. In case a^b 2 — a 2 b x = 0, the 
two lines are parallel and the substitution z = a 2 x + b 2 y or z = a x x + b x y will 
separate the variables. 

7. By the method of Ex. 6 solve the equations : 

(a) (3y-7x + 7)dx + (7y-3x + 3)dy = 0, -Ans. (y - x + l) 2 (y + x- l) 7 = C. 

(/3) (2x + 3y-5)y' + (3x + 2y-5)=0, (7) (4x+3y+l)dx+ (x + y + l)d</=0, 

dy I x — y — 1 



(5) (2x + y) = y'(4x+2y-l), (e) / = (-^— 

dx \z x — 



2y+l 



8. Show that if the equation may be written as yf(xy)dx-\-xg(xy)dy = 0, 
where / and g are functions of the product xy, the substitution v = xy will sepa- 
rate the variables. 

9. By virtue of Ex. 8 integrate the equations : 

(a) (y + 2xy 2 — x 2 y 3 ) dx + 2 x 2 ydy - 0, Ans. x + x 2 y = C (1 — xy). 

(P) (y + xy 2 ) dx + (x - x 2 y) dy = 0, (7) (1 + xy) xyHx + (xy - 1) xdy = 0. 



COMMONER ORDINARY EQUATIONS 207 

10. By any method that is applicable solve the following. If more than one 
method is applicable, state what methods, and any apparent reasons for choos- 
ing one : 

(a) y' + y cos x — y n sin 2x, (/3) (2 x 2 y + 3 y 3 ) dx = (x 3 + 2 xy 2 ) dy, 

(y) (4x + 2y-l)y' + 2x+y-|-l = 0, (5) y y' + xy 2 = x, 

(e) y' sin'y + sin x cos y = sin x, (f) Va 2 + a: 2 (l — y') = x + y, 

(77) (x 3 y 3 + x 2 y 2 + xy + 1) y + (x 3 y 3 — x 2 y 2 — xy + 1) xy', (5) y' = sin (x — y), 

( i ) xydy — y 2 dx = (x + y) 2 e * dx, (k) (1 — y 2 ) dx = axy (x + 1) dy. 

91. Integrating factors. If the equation Mdx + 2V<&/ = by a suita- 
ble rearrangement of the terms can be put in the form of a sum of total 
differentials of certain functions u, v, • • • , say 

du + dv H = 0, then m+H = C (7) 

is surely the solution of the equation. In this case the equation is called 
an exact differential equation. It frequently happens that although the 
equation cannot itself be so arranged, yet the equation obtained from 
it by multiplying through with a certain factor /jl(x, y) may be so 
arranged. The factor fi(x f y) is then called an integrating factor of the 
given equation. Thus in the case of variables separable, an integrating 
factor is l/MJSf x ; for 

-^[.V r V 2 ^ + ^ 2( /,] = |M^ + |g^ = 0; (8) 

and the integration is immediate. Again, the linear equation may be 
treated by an integrating factor. Let 

dy -h X x ydx = X 2 dx and fi = e f Xldx ; (9) 

then ef^ dx dy + X x eS^ dx ydx = ef^ dx X 2 dx (10) 

or d\_yeJ^ dx \ = eI x ^ dx X 2 dx, and yef* dx = J ef x * dx X 2 dx. (11) 

In the case of variables separable the use of an integrating factor is 
therefore implied in the process of separating the variables. In the 
case of the linear equation the use of the integrating factor is somewhat 
shorter than the use of the substitution for separating the variables. 
In general it is not possible to hit upon an integrating factor by inspec- 
tion and not practicable to obtain an integrating factor by analysis, but 
the integration of an equation is so simple when the factor is known, 
and the equations which arise in practice so frequently do have simple 
integrating factors, that it is worth while to examine the equation to 
see if the factor cannot be determined by inspection and trial. To aid 
in the work, the differentials of the simpler functions such as 



208 DIFFERENTIAL EQUATIONS 

dxy = xdy + ydx, ^ d (x 2 ■+- y 2 ) = xdx + ydy, 

■ y xdy — ydx , , -x i/dx — xdy 

d+= J 2 J > dtan" 1 - = ^- — r-2, 

jc x* y ar + 2/ 

should be borne in mind. 



(12) 



Consider the equation {x^e x — 2 rax?/ 2 ) dx + 2 mx 2 ydy = 0. Here the first term 
jc^cZx will be a differential of a function of x no matter what function of x may be 
assumed as a trial n. With fi = 1/x 4 the equation takes the form 

\ X 2 X 3 ) X 1 

The integral is therefore seen to be &° + my 2 /x 2 = C without more ado. It may- 
be noticed that this equation is of the Bernoulli type and that an integration by 
that method would be considerably longer and more tedious than this use of an 
integrating factor. 

Again, consider (x + y)dx — (x — y)dy — and let it be written as 

xdx + ydy + ydx — xdy = ; try n = l/(x 2 + y 2 ) ; 

xdx + ydy ydx — xdy n 1 , . , „ _. , , x 

then , J r— 1 r- = or -d log x 2 + y 2 ) + d tan-i - = 0, 

x 2 + y 2 x 2 + y 2 2 y 



and the integral is log V x 2 + y 2 + tan -1 (x/y) = C Here the terms xdx + ydy 
strongly suggested x 2 + y 2 and the known form of the differential of tan- 1 (x/y) 
corroborated the idea. This equation comes under the homogeneous type, but the 
use of the integrating factor considerably shortens the work of integration. 

92. The attempt has been to write Mdx + Ndy or //, (Mdx + Ndy) 
as the sum of total differentials du -f dv + • • • , that is, as the differential 

dF of the function u-\-v~-\ , so that the solution of the equation 

Mdx + Ndy = could be obtained as F= C. When the expressions 
are complicated, the attempt may fail in practice even where it theoreti- 
cally should succeed. It is therefore of importance to establish condi- 
tions under which a differential expression like Pdx + Qdy shall be the 
total differential dF of some function, and to find a means of obtaining 
F when the conditions are satisfied. This will now be done. 

cF dF 

Suppose Pdx + Qdy = dF = — dx + — dy ; (13) 

„ dF dF dP dQ d 2 F 

then p = —-, q— , == ^^ . 

ex vy cy ox excy 

Hence if Pdx -f- Qdy is a total differential dF, it follows (as in § 52) that 
the relation P' y = Q' x must hold. Now conversely if this relation does 
hold, it may be shown that Pdx -f Qdy is the total differential of a 
function, and that this function is 



COMMONER ORDINARY EQUATIONS 209 

F= f P(x,y)dx + fQ(x Q ,y)dy 

° (14) 

Q(x,ij)dy+ I P(x,y )dx, 
y J 

where the fixed value x or y will naturally be so chosen as to simplify 
the integrations as much as possible. 

To show that these expressions may be taken as F it is merely neces- 
sary to compute their derivatives for identification with P and Q. Now 

dF d r x d r 

= Yx ) P ' X ' y ) dx + di) Q ( x °' y > dy = P ( x ' y }' 



dx 
dF 



= dy j P ( X ' y ^ dX + dy\ Q ( X * y ^ dy== dyJ PdX + Q ^ ' ^' 



These differentiations, applied to the first form of F, require only the 
fact that the derivative of an integral is the integrand. The first turns 
out satisfactorily. The second must be simplified by interchanging the 
order of differentiation by y and integration by x (Leibniz's Rule,. 
§ 119) and by use of the fundamental hypothesis that P , y =Q' x . 



c r x r x dp 

I Pdx + Q(x , y)= I JZ dx + Q ( x o> V) 

J x n Jx n J 



C !J 



-£ 



\x 



j£dx + Q(x Q , y)=Q(x, y)\ 



,x 

The identity of P and Q with the derivatives of F is therefore estab- 
lished. The second form of F would be treated similarly. 

Show that (x 2 + log y) dx + x/ydy = is an exact differential equation and obtain 
the solution. Here it is first necessary to apply the test P' y = Q x . Xow 

9 . . 1 ex 1 

— (x 2 + log y) = - and ■ = - . 

cy y cxy y 

Hence the test is satisfied and the integral is obtained by applying the formula : 

J^x p o \ 

(x 2 + log y) dx + I - dy = - x 3 + x log y = C 
o J y 3 

-dy+l (x 2 + log 1) dx = x log y + - x 3 = C. 
l y J 3 

It should be noticed that the choice of x = simplifies the integration in the first 
case because the substitution of the lower limit is easy and because the second 
integral vanishes. The choice of y Q = 1 introduces corresponding simplifications in 
the second case. 



210 



DIFFERENTIAL EQUATIONS 



Derive the partial differential equation which any integrating factor of the differ- 
ential equation Mdx + Ndy = must satisfy. If fx is an integrating factor, then 

d/xM _ dfxN 
dy dx 

'dN dM\ 

~dy) 



Hence 



txMdx + ixNdy = dF and 



M**-N** 



(15) 



is the desired equation. To determine the integrating factor by solving this equa- 
tion would in general be as difficult as solving the original equation ; in some 
special cases, however, this equation is useful in determining /x. 

93. It is now convenient to tabulate a list of different types of dif- 
ferential equations for which an integrating factor of a standard form 
can be given. With the knowledge of the factor, the equations may 
then be integrated by (14) or by inspection. 



Equation Mdx -f- Ndy = : 
I. Homogeneous Mdx + Ndy == 0, 

II. Bernoulli dy + X^dx = X 2 y n dx, 
III. M = yf(xy), N = xg(xy), 

d_M d_N . ■ 
dy dx 



Factor /j, 
1 



Mx -f- 


Ny 


y~n e a- 

1 


..)/. 



IV. If 



V. If 



N 




dN 
dx 


dM 



M 



■m, 



= /:(£)< 



VI. Type x a y B (mydx + nxdy) = 0, 
VII. x a y 3 (mydx + nxdy) -f- xyy s (pydx -f- qxdy) = 0, 



Mx — Ny 

e ff(x)dx^ 
s ~,km — 1— aykn — 1— 18 

lk arbitrary. 

s -Jem —l — a„.kn—l—fi 

lk determined.- 



The use of the integrating factor often is simpler than the substitu- 
tion y = vx in the homogeneous equation. It is practically identical 
with the substitution in the Bernoulli type. In the third type it is 
often shorter than the substitution. The remaining types have had no 
substitution indicated for them. The proofs that the assigned forms 
of the factor are right are given in the examples below or are left as 
exercises. 

To show that n — (Mx + Ny)- 1 is an integrating factor for the homogeneous 
case, it is possible simply to substitute in the equation (15), which jx must satisfy, 
and show that the equation actually holds by virtue of the fact that M and N are. 



COMMONER ORDINARY EQUATIONS 211 

homogeneous of the. same degree, — this fact being used to simplify the result by 
Euler's Formula (30) of § 53. But it is easier to proceed directly to show 

1 * =±( - V ) or A/LJ_\ = »(1_JL_Y where * = **. 
cy Mx + Ny ex \Mx + Ny/ dy\xl-\- <f>/ ex \y 1 + <f>/ Mx 

Owing to the homogeneity, is a function of ij/x alone. Differentiate. 

C /l 1 \ 1 / 1 = 1 / -y = g A 

cy \x 1 + 0/ X (1 + 0) 2 X y (1 + 0) 2 x 2 CX \?/ 1 + 

As this is an evident identity, the theorem is proved. 

To find the condition that the integrating factor may be a function of x only 
and to find the factor when the condition is satisfied, the equation (15) which /x 
satisfies may be put in the more compact form by dividing by fx. 

M-^-N-^=—-— or M dl ° Sfl Jsr dlogfi = dN dM . (15') 
li cy fidx ex cy cy ex ex cy 

Now if ix (and hence log fx) is a function of x alone, the first term vanishes and 

aiogn K~ N . 



dx 



f(x) or log/* = ff(x) 



dx . 



This establishes the rule of type IV above and further shows that in no other case 
can fx be a function of x alone. The treatment of type V is clearly analogous. 

Integrate the equation x 4 ?/ (3 ydx + 2 xdy) + x 2 (4 ydx + 3 xdy) = 0. This is of 
type VII ; an integrating factor of the form /x = xPy* will be assumed and the ex- 
ponents /o, 0- will be determined so as to satisfy the condition that the equation be 
an exact differential. Here 

P = fxM = 3 xP + *ij* + 2 + 4 xP + 2 2/°" + 1 , Q = fiN = 2 xp + hj* +i + 3 xp + Hj*. 

Then P' y = 3 (<r + 2)xP + 4 ^+ 1 + 4(<r'+ lJasP + V 

= 2(/d + 5)xP + 4 ^+i+3(/d + 3)xP + 2 ^= Q' x . 

Hence if 3 (<r + 2) = 2 (p + 5) and 4 (<r + 1) = 3 (p + 3), 

the relation P^ = Q x will hold. This gives <r = 2, p = 1. Hence /x — xy' 2 , 



and 



f X (3x 5 ?/ 4 + 4x-V 3 )^x + fody = *x 6 ?/ 4 + xV = C 



is the solution. The work might be shortened a trifle by dividing through in the 
first place by x 2 . Moreover the integration can be performed at sight without the 
use of (14). 

94. Several of the most important facts relative to integrating factors 
and solutions of Mdx -f- Ndy = will now be stated as theorems and 
the proofs will be indicated below. 

1. If an integrating factor is known, the corresponding solution may 
be found ; and conversely if the solution is known, the corresponding 
integrating factor may be found. Hence the existence of either implies 
the existence of the other. 

2. If F = C and G — C are two solutions of the equation, either must 
be a function of the other, as G = &(F) ; and any function of either is 



212 DIFFERENTIAL EQUATIONS 

a solution. If /jl and v are two integrating factors of the equation, the 
ratio fi/v is either constant or a solution of the equation ; and the prod- 
uct of fji by any function of a solution, as /jl&(F), is an integrating fac- 
tor of the equation. 

3. The normal derivative dF/dn of a solution obtained from the 
factor fju is the product \i Vilf 2 + N 2 (see § 48). 

It has already been seen that if an integrating factor /x is known, the corre- 
sponding solution F = C may be found by (14). Now if the solution is known, the 
equation 

dF = F x dx + F' y dy = fx (Mdx + Ndtj) gives F' x = fiM, F y = fxN ; 

and hence /x may be found from either of these equations as the quotient of a 
derivative of F by a coefficient of the differential equation. The statement 1 is 
therefore proved. It may be remarked that the discussion of approximate solutions 
to differential equations (§§ 86-88), combined with the theory of limits (beyond the 
scope of this text), affords a demonstration that any equation Mdx -\- Ndy = 0, 
where M and N satisfy certain restrictive conditions, has a solution ; and hence it 
may be inferred that such an equation has an integrating factor. 

If ix be eliminated from the relations F x = /xM, F' y = /xN found above, it is seen 

that 

MF' y - NF X = 0, and similarly, MG y - NG X = 0, (16) 

are the conditions that F and G should be solutions of the differential equation. 
Now these are two simultaneous homogeneous equations of the first degree in M 
and N. If M and N are eliminated from them, there results the equation 



kg: - f'g: = o 



F' F' 

g: g: 



J(F, G) = 0, (16') 



which shows (§ 62) that F and G are functionally related as required. To show 
that any function 3> (F) is a solution, consider the equation 

m; - n*' x = (mf; -nf x ) v. 

As F is a solution, the expression MF' y —NF x vanishes by (16), and hence M$ y —N$' x 
also vanishes, and * is a solution of the equation as is desired. The first half of 2 
is proved. 

Next, if fx and v are two integrating factors, equation (IS") gives 

M dl ° glM x dl ° Sf * = M dl ° 8V N dl0gv or M dl ° gfi/v N 8log,l/p = 0. 
dy dx dy dx dy dx . 

On comparing with (16) it then appears that log (/x/p) must be a solution of the 
equation and hence /x/v itself must be a solution. The inference, however, would 
not hold if /x/v reduced to a constant. Finally if /x is an integrating factor leading 
to the solution F = C, then 

dF = (x (Mdx + Ndy), and hence ^ {F) (Mdx + Ndy) = d f<i> (F) dF. 

It therefore appears that the factor /*<£ (F) makes the equation an exact differen- 
tial and must be an integrating factor. Statement 2 is therefore wholly proved. 



COMMONER ORDINARY EQUATIONS 213 

The third proposition is proved simply by differentiation and substitution. For 

dF _BFdx dF dy _ dx dy 

dn ex dn dy dn dn dn 

And if r denotes the inclination of the curve F = C, it follows that 

dy M . dy N dx M 

tan t = — - = , sm t = — = — , , — cos t — 



dx N dn V-M" 2 + N 2 dn Vilf 2 4- N 2 



Hence dF/dn = fx V M 2 + N 2 and the proposition is proved. 

EXERCISES 

1. Find the integrating factor by inspection and integrate : 

(a) xdy — ydx = (x 2 + y 2 ) dx, (/3) (y 2 — xy) dx + x 2 dy = 0, 

(7) ydx — xdy + logxdx = 0, (5) y (2 xy + e x ) dx — e?dy = 0, 

( e) (1 + xy) ydx + (1 - xy) xdy = 0, (f ) (x - y 2 ) dx + 2 xydy = 0, 

(y) (xy 2 + y)dx — xdy = 0, (6) a (xdy + 2 ydx) = xydy, 

(l) (x 2 + y 2 ) (xdx + ydy) + Vl + (x 2 + y 2 ) (ydx - xdy) = 0, 

(k) x 2 ydx — (x 3 + y s )dy = 0, (X) xdy — ydx = x Vx 2 — y 2 dy. 

2. Integrate these linear equations with an integrating factor : 

(a) y' + ay = sin bx, (£) ?/' + y cot x = sec x, 

(7) (x+ l)2/ / -2?/ = (x+ l) 4 , (8) (l+x 2 )y' + y = et™- 1 *, 

and (/3), (5), (f) of Ex. 4, p. 206. 

3. Show that the expression given under II, p. 210, is an integrating factor for 
the Bernoulli equation, and integrate the. following equations by that method : 

(a) y' — y tan x = y 4 sec x, (j3) 3 y 2 y' + y s = x — 1, 

(7) y" + V cos x — y n sin 2 x, (5) dx + 2 xydy = 2 ax s y 3 dy, 

and (a), (7), (e), (7?) of Ex. 4, p. 206. 

4. Show the following are exact differential equations and integrate : 

(a) (3x 2 + 6xij 2 )dx+(6x 2 y+4:y 2 )dy = 0, (/3) sin x cos ydx + cos x sin ydy = 0, 
(7) (6x-2y + l) + (22/-2x-3)d 2 / = 0, (5) (x 3 + 3 xy 2 ) dx + (y 3 + 3 x 2 y) dy = 0, 

( e ) 2 ^ + 1 dx + ^^#z,0, (f) (l+^)dx+e^l--\d?/ = 0. 

y 2/- V y! 

(ri) e x (x 2 + y 2 + 2 x) dx + 2 ye*d?/ = 0, (0) (y sin x — 1) dx + (?/ — cos x) dy - 0. 

5. Show that (Mx — JSfy)- 1 is an integrating factor for type III. Determine 
the integrating factors of the following equations, thus render them exact, and 
integrate : 

(a) (y + x) dx + xdy = 0, (/3) (?/ 2 - xy) dx + x 2 dy = 0, 

(7) (x 2 +_y 2 ) dx - 2 xydy = 0, (5) (x 2 y 2 + xy) ydx + (x 2 ?/ - 1) xdy = 0, 

( e ) ( -Vxy - l) xdy - (Vxy + l) ydx = 0, ( f ) x 3 dx + (3 x 2 y + 2 y 3 ) dy = 0, 

and Exs. 3 and 9, p. 206. 

6. Show that the factor given for type VI is right, and that the form given for 
type VII is right if k satisfies k (qm — pn) = q (a — 7) — p (/3 — 5). 



214 DIFFERENTIAL EQUATIONS 

7. Integrate the following equations of types IV-VII : 

(a) (z/ 4 + 2 y) dx + (xy* + 2y*-4:x)dy = 0, (/3) (x 2 + y* + 1) dx - 2 zydy = 0, 
( 7 ) (3x 2 + Gxy3y 2 )dx+ (2x* + 3xy)dy = 0, (5) (2xV 2 + y)- [xhj - Sx)y' = 0, 

(e) (2x 2 y-3?/ 4 )dx + (3x 3 + 2x?/ 3 )cty = 0, 

(f) (2 - y') sin (Sx - 2y) + y' sin (x - 2 y) = 0. 

8. By virtue of proposition 2 above, it follows that if an equation is exact and 
homogeneous, or exact and has the variables separable, or homogeneous and under 
types IV-VII, so that two different integrating factors may be obtained, the solu- 
tion of the equation may be obtained without integration. Apply this to finding 
the solutions of Ex. 4 (/S), (5), (7) ; Ex. 5 (or), (7). 

9. Discuss the apparent exceptions to the rules for types I, III, VII, that is, 
when Mx + Ny = or Mx — Ny = or qm — pn = 0. 

10. Consider this rule for integrating Mdx + Xdy=0 when the equation is known 
to be exact : Integrate Mdx regarding y as constant, differentiate the result regard- 
ing y as variable, and subtract from N ; then integrate the difference with respect 
to y. In symbols, 



C 



= f (Mdx + Ndy) = f Mdx + f (x - — f Mdx\dy. 



Apply this instead of (14) to Ex. 4. Observe that in no case should either this 
formula or (14) be applied when the integral is obtainable by inspection. 

95. Linear equations with constant coefficients. The type 

«og + « 1 £| + ..^«,_ 1 | + ^ = X(x) (17) 

of differential equation of the nth order which is of the first degree in 
y and its derivatives is called a linear equation. For the present only 
the case where the coefficients a , a v ••-, a n _ x , a n are constant will be 
treated, and for convenience it will be assumed that the equation has 
been divided through by d Q so that the coefficient of the highest deriva- 
tive is 1. Then if differentiation be denoted by D, the equation may be 
written symbolically as 

(D» + a x JT - 1 + • • • + a n _,D + a n ) y = X, (17') 

where the symbol D combined with constants follows many of the laws 
of ordinary algebraic quantities (see § TO). 

The simplest equation w r ould be of the first order. Here 



dy 

dx 



y~ — a x y = X and y = e° lX I e~ a ^Xdx, (18) 

as may be seen by reference to (11) or (6). Now if D — a x be treated 
as an algebraic symbol, the solution may be indicated as 

(D-a x )y = X and y = ^±—x, (18') 



COMMONER OEDIXAEY EQUATIONS 215 

where the operator (D — r^) -1 is the inverse of D — « 1 . The solution 
which has just been obtained shows that the interpretation which must 
be assigned to the inverse operator is 

— — — (*) - e «i* I 6 -«i*(#) dx, (19) 

where (*) denotes the function of x upon which it operates. That the 
integrating operator is the inverse of D — a x may be proved by direct 
differentiation (see Ex. 7, p. 152). 

This operational method may at once be extended to obtain the solu- 
tion of equations of higher order. Eor consider 

S + a i^ + ^ = z or W + y> + «tfy*v*. (20) 

Let a 1 and a 2 be the roots of the equation D 2 + a^D + a 2 = so that 
the differential equation may be written in the form 

[D 2 -(« 1 +a 2 )D + a 1 a 2 ] !/ = X or (D - aj (D - aj y = X. (20') 

The solution may now be evaluated by a succession of steps as 

(D -a,)ij = - _ A' = e aiX f e~^ x Xdx, 
y = x = e a * x \ e~ a A e a ' x I e~ a ^ x Xdx 

y = e^Je^ x [je^ x Xdx]dx. (20") 

The solution of the equation is thus reduced to quadratures. 

The extension of the method to an equation of any order is immediate. 
The first step in the solution is to solve the equation 

D n + afi— 1 + ■■,. + a n _J) + a n = 

so that the differential equation may be written in the form 

(D _ ai) (p _ a% ) • • • (D - a n _ x ) (D - a n ) y = X- (17") 

whereupon the solution is comprised in the formula 



y 



= e a * x f ePn-L-«n)x I ... I 6 ^- a *> x f e~ a i x X(dx) n , (17'") 

where the successive integrations are to be performed by beginning 
upon the extreme right and working toward the left. Moreover, it 
appears that if the operators D — a n , D — (X n _ 1 , • • ■ , D — a 2 , D — a 1 were 
successively applied to this value of y, they would undo the work here 



216 DIFFERENTIAL EQUATIONS 

done and lead back to the original equation. As n integrations are 
required, there will occur n arbitrary constants of integration in the 
answer for y. 

As an example consider the equation (D 3 — 4D)y = x 2 . Here the roots of the 
algebraic equation D 3 — 4 D = are 0, 2, — 2, and the solution for y is 

The successive integrations are very simple by means of a table. Then 
Ce 2x x 2 dx = \ x 2 e 2x - \ xe 2x + i e 2 * + C t , 

fe-* x fe 2x x 2 (dx) 2 = C{\x 2 e~ 2x - \xe~ 2x + \e~ 2x -\- C x e-± x )dx 

= - ix 2 e-2*- |e-2H C ie -^+ C 2 , 

2/ = C e 2 xCe-±*Ce 2 xx 2 {dxY = f (- \x 2 - | + C 1 e- 2 «+ C 2 e 2 *)dx 

= ~ A * 8 - ¥ & + C^e- 2 * + C 2 e 2x + C 8 . 

This is the solution. It may be noted that in integrating a term like C 1 e~ 4:X the 
result may be written as C^e -4 *, for the reason that C x is arbitrary anyhow ; and, 
moreover, if the integration had introduced any terms such as 2 e~ 2x , \ e 2x , 5, these 
could be combined with the terms G x e~ 2x , C 2 e 2x , C 3 to simplify the form of 
the results. 

In case the roots are imaginary the procedure is the same. Consider 

d 2 y 



+ y = sin x or (_D 2 + 1) y = sin x or (D + i) ( D — i) y = sin x. 
dx 2 

Then y = sin x = e ix fe~ 2 ix fe ix sin x (dx) 2 , i = V— 1. 

The formula for ( e ax sin bxdx, as given in the tables, is not applicable when 

a 2 + b 2 = 0, as is the case here, because the denominator vanishes. It therefore be- 
comes expedient to write sin x in terms of exponentials. Then 

//i pix p— ix pix g— ix 
g- 2 ix I e ix . (^2 . f or gm x — . 

Now —e ix Ce~ 2ix C(e 2ix -l)(dx) 2 = —e ix C e - 2ix \—e 2ix -x + CAdx 



-e fe — x + — e- 2ix x e~ 2ix + C,e~ 2ix + C 9 

! i L2 z 2 i 4 x 2 J 

+ C x e-^+ C 2 e ix . 



2 

x e'"* + e- ix 



2 2 

z>ix J_ p— ix pix «_ p— ix 

Now (V-- + C 2 e- = (C 2 + C x ) -Jl + ( c 2 _ C r ) i g , . 

Hence this expression may be written as C a cos x + C 2 sinx, and then 

2/ — — i x cos x + C' 1 cos x + C 2 sin x. 

The solution of such equations as these gives excellent opportunity to cultivate the 
art of manipulating trigonometric functions through exponentials (§ 74). 



COMMONER OBDINARY EQUATIONS 217 

96. The general method of solution given above may be considerably 
simplified in case the function X(x) has certain special forms. In the 
first place suppose X = 0, and let the equation be P(D)y = 0, where 
P (D) denotes the symbolic polynomial of the nth. degree in D. Suppose 
the roots of P(D) = are a v a 2 , • • • , a k and their respective multiplicities 
are m v m 2 , • • • , m k , so that 

(D — a k ) m k ■ ■ . (D — a 2 ) m *(D — a^p-y = 

is the form of the differential equation. Now, as above, if 

(D-a i r>y = 0, then y = — _* = e^ f ' f<> W" 1 - 

Hence y = e a ^ x (C l + C 2 x -f- C s x 2 -\ f- C m x m ^ - 1 ) 

is annihilated by the application of the operator (D — a^) 1 " 1 , and there- 
fore by the application of the whole operator P (D), and must be a solu- 
tion of the equation. As the factors in P(D) may be written so that 
any one of them, as (D — a>) m i, comes last, it follows that to each factor 
(D — a>) m i will correspond a solution 

Vi = e a i x (C a + C, 2 x +■..-■+ dm^i- 1 ), P (D) Vi = 0, 

of the equation. Moreover the sum of all these solutions, 



y = ]£ e a i*(C a + C i2 x + • • • + dm^i- 1 ), (21) 

i=i 

will be a solution of the equation ; for in applying P{D) to y, 
P(D)y = P(D)y l .+ P(p)y i + ..- + P(p)y t =0. 

Hence the general rule may be stated that : The solution of the dif- 
ferential equation P (D) y = of the nth order 'may be found by multiply- 
ing each e ax by a polynomial of the mth degree in x (where a is a root of 
the equation P (Z>) = of multiplicity m and where the coefficients of the 
polynomial are arbitrary) and adding the results. Two observations 
may be made. Eirst, the solution thus found contains n arbitrary con- 
stants and may therefore be considered as the general solution ; and 
second, if there are imaginary roots for P (D) = 0, the exponentials aris- 
ing from the pure imaginary parts of the roots may be converted into 
trigonometric functions. 

As an example take (D 4 + 2D 3 + D 2 ) y = 0. The roots are 1, 1, 0, 0. Hence the 

solution is ,,//-»'. "-n v. //V.-/-»\ 

y = eP(C 1 + C 2 x) + (C 3 + C 4 x). 

Again if (D 4 + 4) y = 0, the roots of D 4 + 4 = are ± 1 ± i and the solution is 
y = c ie a+0*+ C 2 e(i-o* + C 3 e(- 1+*>*+ C^eC- 1 "^ 



218 DIFFERENTIAL EQUATIONS 

or y = e x (C x e^ + C 2 e~ ix ) + e~ x (C 3 e ix + C 4 e~ ix ) 

= e x (C 1 cos x + C 2 sin x) + e~ x (C 3 cos x + C 4 sin x), 

where the new Cs are not identical with the old O's. Another form is 

?/ = e x ^L cos (x + 7) + e~ x U cos (x + 5), 

where 7 and 5, A and 5, are arbitrary constants. For 

C, cos x + C 2 sin x = V^T^i [^=| ».» + ^=±=** *] . 

if 7= tan- 1 / ?), then C x cos x + C 2 sin x = Vc 2 + C 2 2 cos (x + 7). 



and 



Next if x is not zero but if any one solution I can be found so that 
P(JT) I = X, then a solution containing n arbitrary constants may be 
found by adding to I the solution of P(D)y = 0. For if 

P(D) I = X and P(D) y = 0, then P(D) (/ + y) = X. 

It therefore remains to devise means for finding one solution /. This 
solution / may be found by the long method of (IT'"), where the inte- 
gration may be shortened by omitting the constants of integration since 
only one, and not the general, value of the solution is needed. In the 
most important cases which arise in practice there are, however, some 
very short cuts to the solution /. The solution I of P(D)y = X is 
called the particular integral of the equation and the general solu- 
tion of P(D) y = is called the complementary function for the equa- 
tion P(D) y = X. 

Suppose that X is a polynomial in x. Solve symbolically, arrange 
P (D) in ascending powers of D, and divide out to powers of D equal to 
the order of the polynomial X. Then 



P(D)I=X, 1= — — X=\ Q(D) + 



m*-[«» + ffi. 



X, (22) 



where the remainder R(D) is of higher order in D than X in x. Then 

P(D)I = P(D)Q(D)X + R(D)X, R(D)X=0. 

Hence Q (D) x may be taken as /, since P (D) Q (D) X = P(D)I = X. By 
this method the solution / may be found, when A is a polynomial, as 
rapidly as P (D) can be divided into 1 ; the solution of P (D) y = may 
be. written down by (21) ; and the sum of I and this will be the required 
solution of P (D) y = X containing n constants. 

As an example consider (D 3 + 4D 2 + 3D)y = x 2 . The work is as follows : 

I= 1 X , T= 1 1 ^ = 1 P S 1 18 J> 2 I R(I)) lx 2 

3D+4D 2 + D 3 Z>3 + 4D + D 2 Z>U 9 27 P(D)J 



COMMONER ORDINARY EQUATIONS 219 

26 



— x. 

27 



= Ce c 



Hence 1= Q(D)x- Ig- *2> + |?P 2 ) ^ = ** -^ + 

For P 3 + 4P 2 + 3P = the roots are 0,-1,-3 and the complementary function 
or solution of P(D)y = would be C t + C 2 e~ x + C 3 e _3a; . Hence the solution of 
the equation P (D) ?/ = x 2 is 

y=G x + C 2 e-x + C 3 e- 3x + ±x s - f x 2 + ff as. 

It should be noted that in this example P is a factor of P (P) and has been taken out 
before dividing ; this shortens the work. Furthermore note that, in interpreting 
1/P as integration, the constant may be omitted because any one value of I will do. 

97. Next suppose that X = Ce ax . Now De ax = ae ax , D k e ax = a k e ax , 

and P (P) e ax = P (a) e ax ; hence P (P) — - 

C 
But P(D)T = Ce ax , and hence I = — — e ax (23) 

is clearly a solution of the equation, provided a is not a root of P (P) = 0. 
If P (a) = 0, the division by P (a) is impossible and the quest for I has 
to be directed more carefully. Let a be a root of multiplicity m so that 
P (P) = (P - a) m P x {D). Then 

P a (P) (P - a)™7 = Ce% (P - a)™/ = — — e ax , 

and l = -^ r -e a * / ••• / (dx) m = „ * t • (23') 

P^) J J K > P x (a)m\ K > 

Eor in the integration the constants may be omitted. It follows that 
when X = Ce ax , the solution I may be found by direct substitution. 

Now if X broke up into the sum of terms X = X x + X 2 -\ and if 

solutions Zj, I 2 , • • • were determined for each of the equations P(D)I = X v 
P(D)I 2 = A' 2 , • • ., the solution I corresponding to X would be the sum 

I 1 + I H . Thus it is seen that the above short methods apply to 

equations in which X is a sum of terms of the form Cx m or Ce ax . 

As an example consider (P 4 — 2P 2 + l)y — e x . The roots are 1, 1, — 1, — 1, 
and a = 1 . Hence the solution for I is written as 

(D+l) 2 (P-l) 2 I=e* (P-l) 2 I = ie*, I = \e*x*. 

Then y = e x (G 1 + C 2 x) + e~ x (C 3 + C 4 x) + } e*x 2 . 

Again consider (P 2 — 5D + 6)y = x+ e 7 ^. To find the I x corresponding to x, 

diYkle - , 1 /I 5 „ \ 1 5 

x 



6 36 / 6 



1 6- 5P + P 2 \6 36 / 6 36 
To find the I 2 corresponding to e™*, substitute. There are three cases, 

2 m?-5m+6 2 '2 



220 DIFFERENTIAL EQUATIONS 

according as m is neither 2 nor 3, or is 3, or is 2. Hence for the complete solution, 

15 1 

y = C x e* x + C 2 e 2x + -x + — + 



36 m 2 - 5 m + 6 
when m is neither 2 nor 3 ; tout in these special cases the results are 
y = C x e Sx + C 2 e 2x + ix + I 5 6-a;e 2a; , y = C l e Sx + C 2 e 2x + \x + ^ + ze 3 *. 

The next case to consider is where X is of the form cos fix or sin fix. 
If these trigonometric functions be expressed in terms of exponentials, 
the solution may be conducted by the method above ; and this is per- 
haps the best method when ± fit are roots of the equation P (Z>) = 0. 
It may be noted that this method would apply also to the case where 
X might be of the form e ax cos fix or e ax sin fix. Instead of splitting the 
trigonometric functions into two exponentials, it is possible to combine 
two trigonometric functions into an exponential. Thus, consider the 

equations 

P(D)y = e ax cos fix, P(D)y = e ax sin fix, 

and P(D)y = e ax (cos fix -f i sin fix) = e (a + $ i)x . (24) 

The solution / of this last equation may be found and split into its 
real and imaginary parts, of which the real part is the solution of the 
equation involving the cosine, and the imaginary part the sine. 

When X has the form cos fix or sin fix and ± fit are not roots of the 
equation P(D) = 0, there is a very short method of finding I. For 

D 2 cos fix = — fi 2 cos fix and D 2 sin fix = — fi 2 sin fix. 

Hence if P(B) be written as P X (D 2 ) -f- DP 2 (D 2 ) by collecting the even 
terms and the odd terms so that P 1 and P 2 are both even in D, the 
solution may be carried out symbolically as 

r _ 1 1 1 

■P(P) C ° S X " P r (D 2 ) + DP 2 (D 2 ) C ° b X ~ P x (- fi 2 ) + DP t (- fi 2 ) C ° S *' 

'"[^(-^T + ^C^-^)] 2 ( } 

By this device of substitution and of rationalization as if D were a surd, 
the differentiation is transferred to the numerator and can be performed. 
This method of procedure may be justified directly, or it may be made 
to depend upon that of the paragraph above. 

Consider the example (D 2 + l)y = cosx. Here (3i = i is a root of D 2 + 1 = 0. 
As an operator D 2 is equivalent to — 1, and the rationalization method will not 
work. If the first solution toe followed, the method of solution is 

1 e ix 1 e~ ix 1 e ix 1 e~ ix 1 _ . 1 . 

I = 1 = — = — \xe lx — xe- lx ] = - x sin x. 

D 2 +12D 2 + 12 D-i4ti D + i 4i 4 i l J 2 

If the second suggestion be followed, the solution may toe found as follows : 



COMMONER ORDINARY EQUATIONS 221 





(1)2+1)1 = 


cosx 


+ i sin x = e lx , 


1 = 


1 . xe^ 

giX 

2) 2 + 1 2 i 


Now 


2i 


(cosx 


+ i sin x) = - x sin x - 


1 . 

IX cosx. 

2 


Hence 




1 = 


: i x sin x for 


(2* +1)1 = 


cosx, 


and 




1 = 


— |x cosx for 


P 


2 + l)2 = 


sinx. 


The complete solution is 


y = 


C x cos x + C 2 sin x + 


J x sin x, 




and for 


(2) 2 + 1) y = sin x 


, y = 


C t cosx + C 2 sinx — 


|x cosx. 





As another example take (2) 2 — 3D + 2)y = cosx. The roots are 1, 2, neither 
is equal to ± pi = ± i, and the method of rationalization is practicable. Then 

1 1 1 + 32) 1 ' . . 

Z = cos x = cos x = — cos x = — (cos x — 3 sin x) . 

2) 2 -32)+2 1-32) 10 10 V ' 

The complete solution is y = C^e~ x + C 2 e-' 2x + T L(cosx — 3 sinx). The extreme 
simplicity of this substitution-rationalization method is noteworthy. 



EXERCISES 

1. By the general method solve the equations : 

dx 2 dx dx z dx 2 dx 

( 7 ) (2) 2 -42)+2)?/ = x, (5) (2>HD 2 -4D+4)!/ = a;, 

(e) (D* + 52) 2 + 6 2))y = x, (f) (2) 2 + 2) + 1) y = xe*, 

( v ) (2) 2 + 2)+ l)?y = sin2x, (6) (D 2 - 4)y = x + e 2 *, 

(i) (2) 2 + 3 2) + 2) y = x + cosx, (/c) (D 4 - 42) 2 ) y = 1 - sin x, 

(X ) (2) 2 + 1) y = cos x, (/*) (2) 2 + 1) y = sec x, (?) (D 2 + l)y = tan x. 

2. By the rule write the solutions of these equations : 

{a) (2) 2 + 3 2) + 2) y = 0, (p) (2) 3 + 3 1) 2 + D - 5) y = 0, 

( 7 ) (2)-l) 3 y = 0, (5) (2) 4 + 22)2 + l)y = 0, 

(e) (2) 3 -32) 2 + 4)y = 0, (f) (2) 4 - 2) 3 - 9 2) 2 - 11 D- 4)y = 0, 

(77) (2) 3 - 6 D 2 + 9 D) y = 0, (0) (D 4 - 42) 3 + 82) 2 - 82) + 4) y = 0, 

(t) (2> 5 -22) 4 + 2)8) y = 0, (/c) (2) 3 -2) 2 + 2))y = 0, 

(X) (2) 4 - l) 2 y = 0, (/*) (2)5 _ 132) 3 + 26 2) 2 + 822) + 104)2/ = 0. 

3. By the short method solve (7), (5), (e) of Ex. 1, and also : 

(a) (2) 4 - 1) y = x 4 , (p) (2) 3 -62) 2 +112)-6)y = x, 

(7) (2) 3 + 3 2) 2 + 2 2))y = x 2 , (5) (2) 3 -32) 2 -62) + 8)y = x, 

(e) (2) 3 + 8)y = x 4 + 2x + l, (f) (2) 3 -32) 2 -2)+ 3)y = x 2 , 

( v ) (2> - 2 2) 3 + 2) 2 ) y = x, (0) (2) 4 + 22) 3 + 32) 2 + 22)+l)y = l + x + x 2 , 

( 1 ) (2)3 - 1) */ = x 2 , ( K ) (2) 4 - 2 2) 3 + 2) 2 ) y = x 3 . 

4. By the short method solve (a), (p), (0) of Ex. 1, and also : 

(a) (2) 2 - 3 2) + 2) y = e* (/S) (2) 4 - 2) 3 - 3 2) 2 + 5 2) - 2)y = e 3 *, 

(7) (2) 2 - 2 2) + 1) y = e* (5) (2) 3 - 3 J) 2 + 4) y = e 3 *, 

(e) (2) 2 + l)y = 2e* + x 3 -x, (f) (2) 3 + l)y = 3 + e~* + 5e 2 *, 

(77) (2) 4 + 22)2 + i) y _ e x + ^ (#) (x> 3 + 3 2) 2 + 3 2) + 1) y = 2 e-* 

( t) (2)2-2 2)) y = e 2 * + 1, (k) (2) 3 + 2 2) 2 + 2)) y = e 2x + x 2 + x, 

(X) (2) 2 - a 2 ) y = e«* + e & ^, (p) (2) 2 - 2 a2) + a 2 ) ?/ = e^ + 1. 



222 DIFFERENTIAL EQUATIONS 

5. Solve by the short method (77), (1), (k) of Ex. 1, and also : 

(a) (D 2 - D - 2) y = sin x, (/3) (D 2 + 2 D + 1) = 3 e 2x - cos x, 

(7) (D 2 + 4)y = x 2 + cosx, (5) (D 3 + D 2 - D - l)y = cos2x, 

(e) (D 2 + l) 2 y = cosx, (f) (D 3 - D 2 + D— l)y = cosx, 

(\) (D 2 - 5Z>+ 6)y = cosx- e 2 *, (0) (D 3 - 2i> 2 - 3D)y = 3x 2 + sinx, 

( 1 ) (D 2 - l) 2 y = sin x, (k) (D 2 + 3 D + 2) y = e 2 * sin x, 

(X) (1)4 _ l) y = e * COS x, (ft) (D 3 - 3 D 2 + 4 D - 2) y = 6* + cosx, 

(v) (I) 2 — 2 D + 4) y = e* sin x, (o) (D 2 + 4) y = sin 3 x + e* + x 2 , 

- xV3 

(7r) (D 6 + 1) y = sin f x sin i x, (p) (D 3 + l)y = e 2x sinx + e 2 sin , 

z 

(a) (D 2 + 4)2/ = sin 2 x, (r) (D 4 + 32 D + 48) y = xe~ 2x + e 2 *cos2f x. 

6. If X has the form e« x X,, show that 1 = — - — e ax X, = e ax 



P(D) T P(D+a) v 

This enables the solution of equations where X 1 is a polynomial to be obtained by 
a short method ; it also gives a way of treating equations where X is e ax cos fix or 
e ax sin /3x, but is not an improvement on (24) ; finally, combined with the second 
suggestion of (24), it covers the- case where X is the product of a sine or cosine by 
a polynomial. Solve by this method, or partly by this method, (f) of Ex. 1 ; (/c), (X), 
(v), (p), (t) of Ex. 5 ; and also 

(a) (D 2 - 2 D + 1) y = x 2 e 3 *, (j3) (D 3 + 3 D 2 + 3 1) + 1) y = (2 - x 2 ) e-« 

( 7 ) (£>2 + 7l 2) y = x 4 e ^ (g) (X)4 _ 2 1)3 _ 3 x>2 + 4 x> + 4) y = x 2 e x , 

(e) (D 3 -7D-6)t/ = e 2a; (l + x), (J) (Z> — l) 2 y = e x + cos x + x 2 e*, 

( v ) (D - l) 3 y = x - x 3 e*, (0) (D 2 + 2)y = x 2 e 3 * + e*cos2x, 

( i ) (i» 3 — 1) y = xe* + cos 2 x, (k) (D 2 - 1) y = x sin x + (1 + x 2 ) e x , 

(X) (D 2 + 4) y = x sin x, (fi) (£> 4 + 2 D 2 + 1) y = x 2 cos ax,_ 

(v) {I) 2 + 4) y = (x sin x) 2 , (o) (D 2 — 2 D + 4) 2 y = xe* cos V3 x. 

7. Show that the substitution x = e l , Ex. 9, p. 152, changes equations of the type 

x n D n y + a x x n -^D n ~Hj + • • • + a w _ixDy + a n y = X(x) (26) 

into equations with constant coefficients ; also that ax -{- b = e t would make a simi- 
lar simplification for equations whose coefficients were powers of ax + b. Hence 
integrate : 

(a) (x 2 I) 2 - xD + 2) y = x log x, (j8) (x 3 D 3 - x 2 D 2 + 2 xD - 2) y = x 3 + 3 x, 

(7) [(2x-l) 3 D 3 +(2x-l)Z>-2]?/ = 0, (5) (x 2 D 2 + 3xD + l)y = (1 - x)~ 2 , 
(e) (x 3 D 3 + xD - 1) y = x log x, (f) [(x + 1) 2 D 2 - 4 (x + 1)D + 6] y = x, 

(77) (x 2 i>+ 4xX> + 2)y = e* (^) (x 3 X> 2 -3x 2 D + x)y= logxsinlogx +1, 

( 1) (x 4 D 4 + 6x 3 D 3 + 4 x 2 £> 2 - 2 xD - 4) y = x 2 + 2 cos log x. 

8. If X be self-induction, R resistance, C capacity, i current, q charge upon the 
plates of a condenser, and/(£) the electromotive force, then the differential equa- 
tions for the circuit are 

, N d 2 q Bdq q 1 ^ v , x dH R di i 1 >//A 

Solve (a) when/(i) = e~ ^ sin hx and (/3) when/(i) = sin bx. Reduce the trigonometric 
part of the particular solution to the form K sin (6x + 7). Show that if R is small 
and b is nearly equal to 1/V-LC, the amplitude K is large. 



COMMONER ORDINARY EQUATIONS 223 

98. Simultaneous linear equations with constant coefficients. If 

there be given two (or in general n) linear equations with constant 
coefficients in two (or in general n) dependent variables and one inde- 
pendent variable t, the symbolic method of solution may still be used 
to advantage. Let the equations be 

(a D* + o^" 1 + ■•• + *,) x + Q>JD m + bJOF-' 1 + ■ • • + b m ) y = R (t), 
(c D* + c x D* - 1 + • • • + c p ) x + (d Q W + d^- 1 + • • • + <2 e ) y = S (*)/ ; 

when there are two variables and where D denotes differentiation by t. 

The equations may also be written more briefly as 

P^D) x + Q X (D) y = R and P,(D) x + Q 2 (D) y == 5. 

The ordinary algebraic process of solution for x and y may be employed 
because it depends only on such laws as are satisfied equally by the 
symbols D, P X (D), Q X (D), and so on. 

Hence the solution for x and y is found by multiplying by the ap- 
propriate coefficients and adding the equations. 

- Q4P) p .(- D ) p l D ) x + U D ) :/ = s - 

Then IPJD) Q. 2 (D) - P„_(D) «,(i>)] x = Q,(D) R - q£d) S, 
lP x (Zf) Q(D) - P 2 (D) Ql (7))] y = Pl (D) S - P 2 (D) R. 

It will be noticed that the coefficients by which the equations are multi- 
plied (written on the left) are so chosen as to make the coefficients of 
x and y in the solved form the same in sign as in other respects. It may 
also be noted that the order of P and Q in the symbolic products is im- 
material. By expanding the operator P a (D) Q (D) — P. 2 ( D ) ^i( D ) a certain 
polynomial in D is obtained and by applying the operators to R and S 
as indicated certain functions of t are obtained. Each equation, whether 
in x or in y, is quite of the form that has been treated in §§ 95-97. 

As an example consider the solution for x and y in the case of 

2^-^_4z = 2*, 2^ + 4&-8i, = 0; 

dt 2 dt dt dt 

or (2 D 2 - 4) x - By = 2 t, 2 JDx + (4 D - 3) y = 0. 

Solve 4 1) - 3 -2D (2D 2 - 4) x - By = 2 « 

D 2 D 2 - 4 2 Da> + (4 D - 3) y = 0. 

Then [(4D - 3) (2D 2 - 4) + 2B 2 ]x =(4D-3)2i, 

[2 D 2 + (2 D 2 - 4) (4D - 3)] y = - (2D)2 «, 
or 4(2D 3 -D 2 -4D+ 3)z = 8- 6«, 4(2 D 3 - D 2 - 4D + S)y = - 4. 

The roots of the polynomial in D are 1, 1, — 1\ ; and the particular solution I x for 
— i Hence the solutions have the form 



(27') 



(C, + Cy) e« +. C 3 e" i« - J t, y = (if, + KJ,) e 



224 DIFFERENTIAL EQUATIONS 

Tlie arbitrary constants which are introduced into the solutions for x 
and y are not independent nor are they identical. The solutions must 
be substituted into one of the equations to establish the necessary relations 
between the constants. It will be noticed that in general the order of the 
equation in D for x and for y is the sum of the orders of the highest 
derivatives which occur in the two equations, — in this case, 3 = 2 -f- 1. 
The order may be diminished by cancellations which occur in the formal 
algebraic solutions for x and y. In fact it is conceivable that the coeffi- 
cient P X Q 2 —Pfii °^ x an( ^- V * n ^ ne SOive( l equations should vanish and 
the solution become illusory. This case is of so little consequence in 
practice that it may be dismissed with the statement that the solution 
is then either impossible or indeterminate ; that is, either there are no 
functions x and y of I which satisfy the two given differential equations, 
or there are an infinite number in each of which other things than the 
constants of integration are arbitrary. 

To finish the example above and determine one set of arbitrary constants in 
terms of the other, substitute in the second differential equation. Then 

2 (C^t + C 2 e* + C 2 tet - f C 3 e~ I * _ i) + 4 (K^ + K 2 e* + K 2 te* - § K 3 e~ t ') 

- 3 (K^ + K 2 te* + K 3 e~ § *■ - \) = 0, 
or e*{2 C X + 2C 2 + E x + K 2 ) + te\2 C 2 + K 2 ) --3e~f \G S + 3 K 3 ) = 0. 

As the terms e', te f , e~2* are independent, the linear relation between them can 
hold only if each of the coefficients vanishes. Hence 

C 3 + 3 K s = 0, 2 C 2 + K 2 = 0, 2 C x + 2 C 2 + K x + K 2 = 0, 

and C 3 = - 3 K 3 , 2C 2 = - K 2 , 2C 1 =-K 1 . 

Hence a?='(C 1 + C 2 t)e*- SK^e'^- \t, y= -2(C 1 + C 2 «)e* + K^e'i'- i 

are the finished solutions, where C x , C 2 , K 3 are three arbitrary constants of inte- 
gration and might equally well be denoted by C x , C 2 , C 3 , or K x , K 2 , K 3 . 

99. One of the most important applications of the theory of simultaneous equa- 
tions with constant coefficients is to the theory of small vibrations about a state of 
equilibrium in a conservative* dynamical system. If q x , q 2 , • • • , q n are n coordinates 
(see Exs. 19-20, p. 112) which specify the position of the system measured relatively 

* The potential energy V is defined as — dV = dW = Q x dq x + Q 2 dq 2 + •■•-(- Q n dq n , 
where to, Sn , to, , to. to, a^ 

Q '- x ' g4i +I 'a ?i +z 'e 5i + --' + x " 8 « / ,. + 1 "e 9i + z »^ 

This is the immediate extension of £q as given in Ex. 19, p. 112. Here dW denotes the 
differential of work and dW = 2 l F i *dr i = 2 (AVtei + !'{<% + Zjtfej). To find Q; it is 
generally quickest to compute d IF from this relation with dX{ , dyi , dzi expressed in terms 
of the differentials dq x , • • • , dq n . The generalized forces Qi are then the coefficients of 
dqi . If there is to be a potential V, the differential d W must be exact. It is frequently 
easy to find V directly in terms of q x , • • • , q n rather than through the mediation of 
Qi > • • ' » Qn ; when this is not so, it is usually better to leave the equations in the form 
d dT cT 

— tt~ — t~ = Qi rather than to introduce V and L. 
dt cq t dqi 



COMMONER ORDINARY EQUATIONS 225 

to a position of stable equilibrium in which all the g's vanish, the development of 
the potential energy by Maclaurin's Formula gives 

V(q 1 , %■> ■ • • , in) = r + Fi^, g 2 , • ■ • , g») + v 2 (q 1 , g 2 , • • • , &) + • • • , 

where the first term is constant, the second is linear, and the third is quadratic, and 
where the supposition that the g's take on only small values, owing to the restriction 
to small vibrations, shows that each term is infinitesimal with respect to the preced- 
ing. Now the constant term may be neglected in any expression of potential energy. 
As the position when all the g's are is assumed to be one of equilibrium, the forces 

Ql — — ~ » 0-2 = 7 ' ' " ' ' Qn = — ~ 

cq x dq 2 cq n 

must all vanish when the g's are 0. This shows that the coefficients, (dV/dqt)o = 0, 
of the linear expression are all zero. Hence the first term in the expansion is the 
quadratic term, and relative to it the higher terms may be disregarded. As the 
position of equilibrium is stable, the system will tend to return to the position 
where all the g's are when it is slightly displaced from that position. It follows 
that the quadratic expression must be definitely positive. 

The kinetic energy is always a quadratic function of the velocities qi , q 2 , • • • , q n 
with coefficients which may be functions of the g's. If each coefficient be expanded 
by the Maclaurin Formula and only the first or constant term be retained, the 
kinetic energy becomes a quadratic function with constant coefficients. Hence the 
Lagrangian function (cf . § 160) 

L=T-V= T(q, , g 2 , . . . , q n ) - V(q lt g 2 , • • • , g.), 

when substituted in the formulas for the motion of the system, gives 

^aL_fi_ ddI L _dL _ ^.^_^_o 

dt cq x cq x dt cq 2 dq 2 ' dt dq n dq n 

a set of equations of the second order with constant coefficients. The equations 
moreover involve the operator D only through its square, and the roots of the equa- 
tion in D must be either real or pure imaginary. The pure imaginary roots intro- 
duce trigonometric functions in the solution and represent vibrations. If there were 
real roots, which would have to occur in pairs, the positive root would represent 
a term of exponential form which would increase indefinitely with the time, — a 
result which is at variance both with the assumption of stable equilibrium and 
with the fact that the energy of the system is constant. 

When there is friction in the system, the forces of friction are supposed to vary 
with the velocities for small vibrations. In this case there exists a dissipative func- 
tion F ( q x , q 2 , • • • , q n ) which is quadratic in the velocities and may be assumed to 
have constant coefficients. The equations of motion of the system then become 

A^_f^± + £^-o d dL dL dF -0 

dt c'q A cq x cq x ' dt cq n cq n cq n 

which are still linear with constant coefficients but involve first powers of the 
operator D. It is physically obvious that the roots of the equation in D must be 
negative if real, and must have their real parts negative if the roots are complex ; 
for otherwise the energy of the motion would increase indefinitely with the time, 
whereas it is known to be steadily dissipating its initial energy. It may be added 
that if, in addition to the internal forces arising from the potential V and the 



226 DIFFERENTIAL EQUATIONS 

frictional forces arising from the dissipative function F, there are other forces 
impressed on the system, these forces would remain to be inserted upon the right- 
hand side of the equations of motion just given. 

The fact that the equations for small vibrations lead to equations with constant 
coefficients by neglecting the higher powers of the variables gives the important 
physical theorem of the superposition of small vibrations. The theorem is : If with 
a certain set of initial conditions, a system executes a certain motion ; and if with 
a different set of initial conditions taken at the same initial time, the system 
executes a second motion ; then the system may execute the motion which consists 
of merely adding or superposing these motions at each instant of time ; and in 
particular this combined motion will be that which the system would execute under 
initial conditions which are found by simply adding the corresponding values in 
the two sets of initial conditions. This theorem is of course a mere corollary of the 
linearity of the equations. 

EXERCISES 

1. Integrate the following systems of equations : 

(a) Dx — Dy + x = cos t, 
(/3) 3Dx + 3x + 2y = &, 
( 7 ) D 2 x- 3x- 4# = 0, 
dx 



(*) 


v — < 


7 X 


2 x + 5 y 


dt, 


(f) Wx + 2(x 


- y) = i; 




(v) 


Dx = 


ny - 


- mz, 




(0) 


D 2 x- 


- Sx'- 


-4y + 3 = 


= 0, 


(0 


D 4 x- 


-4Z> 3 z/ + 4D 2 x 


— X 



D^x-Dij + 3x-y = e*t, 




4x-SDy + Sy = St, 




DHj + x + y = 0, 




(c) dt- dX - dy 




3x+4# 2x+5y 




tDy + x + 5 y = t, 




Dy = lz — ?ix, Dz = mx - 


-iy-. 


D*y + x-8y + 5 = 0, 




D±y - 4 D s x + ±D 2 y-y = 0. 





o, 

2. A particle vibrates without friction upon the inner surface of an ellipsoid. 
Discuss the motion. Take the ellipsoid as 

a 2 6 2 c 2 



then x = CsiiW— ^i+ C 1 ), ?/ = JSTsinf — —t + ^ j. 



3. Same as Ex. 2 when friction varies with the velocity. 

4. Two heavy particles of equal mass are attached to a light string, one at the 
middle, one at one end, and are suspended by attaching the other end of the string 
to a fixed point. If the particles are slightly displaced and the oscillations take 
place without friction in a vertical plane containing the fixed point, discuss the 
motion. 

5. If there be given two electric circuits without capacity, the equations are 

L Jh + M *h + Ri = F L Jh + M S Ri = E 

1 dt dt ll x 2 dt dt 22 2 ' 

where i x , i 2 are the currents in the circuits, L x , L^ are the coefficients of self- 
induction, B x , R 2 are the resistances, and M is the coefficient of mutual induction. 
(a) Integrate the equations when the impressed electromotive forces E 1 , E 2 are 
zero in both circuits. (/3) Also when E 2 = but E^ = sin pt is a periodic force. 
(7) Discuss the cases of loose coupling, that is, where M 2 /L 1 L 2 is small ; and the 
case of close coupling, that is, where M 2 /L l L 2 is nearly unity. What values for p 
are especially noteworthy when the damping is small ? 



COMMOXER OEDIXARY EQUATIONS 227 

6. If the two circuits of Ex. 5 have capacities C x , C 2 and if q x , q 2 are the 
charges on the condensers so that i x = dq x /dt, i 2 = dq 2 /dt are the currents, the 
equations are 

L J^ + M ^ + R ^ + J 1:=E l^ + m^ + eJ^ + ^^e^ 

1 dt°- OP * dt C\ u 2 ffl dP *-dt C 2 2 

Integrate -wh en the resistances are negligible andE 1 = E 2 = 0. If T x = 27r\ / C\L 1 
and T 2 = 2 it V C ' 2 Z 2 are the periods of the i ndividua l separ ate circu its and 
= 2 ttMVc^C^ and if T x = T 2 , show that VJ 2 + e 2 and VjT 2 - e 2 are the 
independent periods in the coupled circuits. 

7. A uniform beam of weight 6 lb. and length 2 ft. is placed orthogonally 
across a rough horizontal cylinder 1 ft. in diameter. To each end of the beam is 
suspended a weight of ] lb. upon a string 1 ft. long. Solve the motion produced 
by giving one of the weights a slight horizontal velocity. Note that in finding the 
kinetic energy of the beam, the beam may be considered as rotating about its 
middle point (§ 39). 



CHAPTER IX 
ADDITIONAL TYPES OF ORDINARY EQUATIONS 

100. Equations of the first order and higher degree. The degree of 
a differential equation is denned as the degree of the derivative of 
highest order which enters in the equation. In the case of the equation 
^r(x, y, y') = of the first order, the degree will be the degree of the 
equation in y'. From the idea of the lineal element (§ 85) it appears 
that if the degree of \l> in y' is n, there will be n lineal elements through 
each point (x, y). Hence it is seen that there are n curves, which are 
compounded of these elements, passing through each point. It may be 
pointed out that equations such as y' = x Vl + y 2 , which are apparently 
of the first degree in y', are really of higher degree if the multiple value 
of the functions, such as Vl + y 2 , which enter in the equation, is taken 
into consideration ; the equation above is replaceable by y' 2 = x 2 -f x 2 y 2 y 
which is of the second degree and without any multiple valued function. # 

First suppose that the differential equation 

* (?, y, y') = W - U x > y)] x [y' - U x > y)] • ■ ■ = ° i 1 ) 

may be solved for y'. It then becomes equivalent to the set 

y' - ^(x, y) = 0, y'- + 2 (x, y) = Q,--- (V) 

of equations each of the first order, and each of these may be treated 
by the methods of Chap. VIII. Thus a set of integrals t 

F 1 (x,y,C) = 0, F 2 (x,y,V) = 0, ... (2) 

may be obtained, and the product of these separate integrals 

F(x, y, C) = Ffa y, C) ■ F 2 (x, y,C) ••■=() (2') 

is the complete solution of the original equation. Geometrically speak- 
ing, each integral F { (x, y, C) = represents a family of curves and the 
product represents all the families simultaneously. 

* It is therefore apparent that the idea of degree as applied in practice is somewhat 
indefinite. 

t The same constant C or any desired function of C may be used in the different 
solutions because C is an arbitrary constant and no specialization is introduced by its 
repeated use in this way. 

228 



ADDITIONAL OEDINAEY TYPES 229 

m 

As an example consider y' 2 + 2 y'y cot x = y 2 . Solve. 

y' 2 + 2 y'y cot x + y 2 cot 2 x = y 2 (l + cot 2 x) = y 2 esc 2 x, 
and (*/' + y cot x — ?/ csc x)(y' + y cot x + 2/ esc x) = 0. 

These equations both come under the type of variables separable. Integrate. 

dy 1 — cos x d cos x 



y sin x 1 + cos x 



2/(1 + cosx) = C, 

, c?2/ 1 + cosx, dcosx , 

and — = dx = , y (1 — cos x) = C. 

y sin x 1 — cos x 

Hence [y (1 + cos x) + C] [y (1 — cos x) + C] = 

is the solution. It may be put in a different form by multiplying out. Then 

y 2 sin 2 x + 2Cy+ C 2 = 0. 

If the equation cannot be solved for y' or if the equations resulting 
from the solution cannot be integrated, this first method fails. In that 
case it may be possible to solve for y or for x and treat the equation by 
differentiation. Let y' =p. Then if 

The equation thus found by differentiation is a differential equation of 
the first order in dp/dx and it may be solved by the methods of Chap. 
VIII to find F(p, x, C) = 0. The two equations 

y =f(x, p) and F(p, x,C)=0 (3') 

may be regarded as defining x and y parametrically in terms of p, or p 
may be eliminated between them to determine the solution in the form 
Q, (x, y, C) = if this is more convenient. If the given differential equa- 
tion had been solved for x, then 

x=f(y,p) and fUl^ + ^fe. (4) 

J w ' 1 J dy p dy Op dy v ) 

The resulting equation on the right is an equation of the first order in 
dp/dy and may be treated in the same way. 

As an example take xp 2 — 2 yp + ax = and solve for y. Then 

ax ^dy dp ax dp a 

2y = xp+—, 2-f- = 2p=p + x^-—-^ + -, 
p dx dx p 2 dx p 

or — I p — — — + ( — — p ) = 0, or xdp —pdx = 0. 



PL pldx \p / 



The solution of this equation is x = Cp. The solution of the given equation is 

2y = xp -\ , x = Cp 

p 

when expressed parametrically in terms of p. lip be eliminated, then 

x 2 

2y = f- aC parabolas. 

G 



230 DIFFEKENTIAL EQUATIONS 

As another example take p 2 y -f 2 px = y and solve for x. Then 

/l \ dx 2 1 , / 1 A dp 

2x = y[--p), 2 — = - = --p + yl --1 -^, 

\p / dy p p \ P 2 / dy 

or - + p + y (— + 1 ) — = 0, or ydp + pdy = 0. 

The solution of this is py = C and the solution of the given equation is 
2 x = y (- - p j , py = C, or y* = 2 Cx + C 2 . 

Two special types of equation may be mentioned in addition, although 
their method of solution is a mere corollary of the methods already 
given in general. They are the equation homogeneous in (x, y) and 
Clairaufs equation. The general form of the homogeneous equation is 
* (P> y/ x ) — 0- This equation may be solved as 

P = t$ oras l = f^> v = xf(p); (5) 

and in the first case is treated by the methods of Chap. VIII, and in 
the second by the methods of this article. Which method is chosen 
rests with the solver. The Clairaut type of equation is 

y=p X +f(p) (6) 

and comes directly under the methods of this article. It is especially 
noteworthy, however, that on differentiating with respect to x the result- 
ing equation is , , 

[*+/(*)] ^ = or ^ = 0. (6') 

Hence the solution for p is p = C, and thus y = Cx -\-f(C) is the solu- 
tion for the Clairaut equation and represents a family of straight lines. 
The rule is merely to substitute C in place of p. This type occurs very 
frequently in geometric applications either directly or in a disguised 4 
form requiring a preliminary change of variable. 

101. To this point the only solution of the differential equation 
\I> (x, y, p) = which has been considered is the general solution 
F(x, y, C)= containing an arbitrary constant. If a special value, 
say 2, is given to C, the solution F(x, y, 2) = is called a particular 
solution. It may happen that the arbitrary constant C enters into the 
expression F(x, y, C) = in such a way that when C becomes positively 
infinite (or negatively infinite) the curve F(x, y, C) = approaches a 
definite limiting position which is a solution of the differential equation ; 
such solutions are called infinite solutions. In addition to these types 
of solution which naturally group themselves in connection with the 
general solution, there is often a solution of a different kind which is 




ADDITIONAL OKDINABY TYPES 231 

known as the singular solution. There are several different definitions 
for the singular solution. That which will be adopted here is : A singu- 
lar solution is the envelope of the family of curves defined by the 
general solution. 

The consideration of the lineal elements (§ 85) will show how it is 
that the envelope (§ 65) of the family of particular solutions which 
constitute the general solution is itself a solution of the equation. For 
consider the figure, which represents the particular solutions broken up 
into their lineal elements. Note that the envelope is made up of those 
lineal elements, one taken from each particular so- 
lution, which are at the points of contact of the ^ envelop e 
envelope with the curves of the family. It is seen jf^ y^R° 
that the envelope is a curve all of whose lineal 
elements satisfy the equation ^ (x, y, pj) = for the 
reason that they lie upon solutions of the equation. Now any curve 
whose lineal elements satisfy the equation is by definition a solution 
of the equation; and so the envelope must be a solution. It might 
conceivably happen that the family F(x, y, C) = was so constituted 
as to envelope one of its own curves. In that case that curve would 
be both a particular and a singular solution. 

If the general solution F(x, y, C) = of a given differential equation 
is known, the singular solution may be found according to the rule for 
finding envelopes (§ 65) by eliminating C from 

F(x, y, C) =0 and ? - F(x, y, C) = 0. (7) 

It should be borne in mind that in the eliminant of these two equations 
there may occur some factors which do not represent envelopes and 
which must be discarded from the singular solution. If only the singu- 
lar solution is desired and the general solution is not known, this 
method is inconvenient. In the case of Clairaut's equation, however, 
where the solution is known, it gives the result immediately as that 
obtained by eliminating C from the two equations 

y = Cx +/(C) and = x +f{C). (8) 

It may be noted that as p = C, the second of the equations is merely 
the factor x +f'(p) = discarded from (6'). The singular solution may 
therefore be found by eliminating p between the given Clairaut equa- 
tion and the discarded factor x -\-f'(p) = 0. 

A reexamination of the figure will suggest a means of finding the 
singular solution without integrating the given equation. For it is seen 
that when two neighboring curves of the family intersect in a point P 



232 DIFFERENTIAL EQUATIONS 

near the envelope, then through this point there are two lineal elements 
which satisfy the differential equation. These two lineal elements have 
nearly the same direction, and indeed the nearer the two neighboring 
curves are to each other the nearer will their intersection lie to the 
envelope and the nearer will the two lineal elements approach coinci- 
dence with each other and with the element upon the envelope at the 
point of contact. Hence for all points (x, y) on the envelope the equa- 
tion \£ (x, y, p) = of the lineal elements must have double roots for p. 
Now if an equation has double roots, the derivative of the equation 
must have a root. Hence the requirement that the two equations 

<A (a, V,P) = and g-r if/ (x, y, p) = (9) 

have a common solution for p will insure that the first has a double 
root for p ; and the points (x, y) which satisfy these equations simul- 
taneously must surely include all the points of the envelope. The rule 
for finding the singular solution is therefore : Eliminate p from the 
given differential equation and its derivative with respect to p, that is, 
from (9). The result should be tested. 

If the equation xp 2 — 2 yp + ax = treated above be tried for a singular solution, 
the elimination of p is required between the two equations 

xp 2 — 2 yp + ax = and xp — y = 0. 

The result is y 2 = ax 2 , which gives a pair of lines through the origin. The substi- 
tution of y = ± Vax and p = ± Va in the given equation shows at once that 
y 2 = ax 2 satisfies the equation. Thus y 2 = ax 2 is a singular solution. The same 
result is found by finding the envelope of the general solution given above. It is 
clear that in this case the singular solution is not a particular solution, as the par- 
ticular solutions are parabolas. 

If the elimination had been carried on by Sylvester's method, then 

x — y 

x — 2 y a = — x (y 2 — ax 2 ) = ; 

x — y 

and the eliminant is the product of two factors x = and y 2 — ax 2 = 0, of which 
the second is that just found and the first is the ?/-axis. As the slope of the ?/-axis 
is infinite, the substitution in the equation is hardly legitimate, and the equation 
can hardly be said to be satisfied. The occurrence of these extraneous factors in 
the eliminant is the real reason for the necessity of testing the result to see if it 
actually represents a singular solution. These extraneous factors may represent 
a great variety of conditions. Thus in the case of the equation p 2 + 2 yp cot x = y 2 
previously treated, the elimination gives y 2 esc 2 x = 0, and as esc x cannot vanish, 
the result reduces to y 2 = 0, or the x-axis. As the slope along the ai-axis is and y 
is 0, the equation is clearly satisfied. Yet the line y = is not the envelope of the 
general solution ; for the curves of the family touch the line only at the points mr. 
It is a particular solution and corresponds to C = 0. There is no singular solution. 



ADDITIONAL OKDINAKY TYPES 233 

Many authors use a great deal of time and space discussing just what may and 
what may not occur among the extraneous loci and how many times it may occur. 
The result is a considerable number of statements which in their details are either 
grossly incomplete or glaringly false or both (cf. §§ 65-67). The rules here given 
for finding singular solutions should not be regarded in any other light than as 
leading to some expressions which are to be examined, the best way one can, to 
find out whether or not they are singular solutions. One curve which may appear in 
the elimination of p and which deserves a note is the tac-locus or locus of points of 
tangency of the particular solutions with each other. Thus in the system of circles 
(x — C) 2 + y 2 = r 2 there may be found two which are tangent to each other at any 
assigned point of the x-axis. This tangency represents two coincident lineal 
elements and hence may be expected to occur in the elimination of p between the 
differential equation of the family and its derivative with respect to p ; but not in 
the eliminant from (7). 

EXERCISES 

1. Integrate the following equations by solving forp = y'\ 

(a) p 2 -6p+5 = 0, (/3) pZ-(2x+ij 2 )p 2 + {x 2 -y 2 + 2xy 2 )p-(x 2 -y 2 )ij 2 = 0, 

(y) xp 2 -2yp-x = 0, {8) p 3 {x + 2y) + Sp 2 (x + y) + p (y + 2 x) = 0, 
(e) y 2 + p 2 = 1, (f) p 2 - ax 3 = 0, ( v ) p = (a - x) Vl + p 2 . 

2. Integrate the following equations by solving for y or x : 

(a) 4 xp 2 + 2 xp — y = 0, (j3) y = — xp + x 4 p 2 , (7) p + 2 xy — x 2 — y 2 = 0, 

(5) 2px — y+ logp = 0, (e) x — yp = ap 2 , (f) y = x + « tan-ip, 

( v ) x = y + a logp, {$) x + py {2p 2 + 3) = 0, ( t ) a 2 yp 2 -2xp + y = 0, 

(k) p 3 — 4xyp + 8 y 2 = 0, (X) x —p + logp, (/*) p 2 (x 2 + 2 ax) = a 2 . 

3. Integrate these equations [substitutions suggested in (t) and (k)] : 

(or) xy 2 {p 2 + 2) =2py* + x 3 ,, (/3) (nx + py) 2 = (1 + P 2 ) (l/ 2 + nx 2 ), 

(7) V 2 + vyP ~ v 2 p 2 = 0, (5) y = yp 2 + 2px, 

(e) y =px+ sin-ip, (f) y = p (x - 6) + a/p, 

(77) y = px +P (1 — P 2 ), (0) y 2 - 2pxy - 1 = p 2 (1 -x 2 ), 

( 1 ) 4 e 2 2/p 2 + 2xp - 1 = 0, z- e 2 v, (x) y = 2px + ?/p 3 , y 2 = z, 

(X) 4e 2 ^p 2 + 2e 2x p~- e x = 0, (/*) x 2 (y — px) = yp 2 . 

4. Treat these equations by the p method (9) to find the singular solutions. 
Also solve and treat by the C method (7). Sketch the family of solutions and 
examine the significance of the extraneous factors as well as that of the factor 
which gives the singular solution : 

(a) p 2 y + p (x — y) — x = 0, (/3) p 2 y 2 cos 2 a — 2pxy sin 2 a + y 2 — x 2 sin 2 a = 0, 

(7) 4xp 2 =: (3 x - a) 2 , ( 5) yp 2 x (x - a) (x - b) = [3x 2 - 2 x (a + b) + a6] 2 , 

(e) p 2 + X p-y = 0, (f) 8a(l + p) 3 = 27(x+?/)(l-p) 3 , 

(i;) x 3 p 2 + x 2 yp + a 3 = 0, (0) 2/(3-4 ?/) 2 p 2 = 4(1 - y). 

5. Examine sundry of the equations of Exs. 1, 2, 3, for singular solutions. 

6. Show that the solution of y = x0(p) +/(p) is given parametrically by the 
given equation and the solution of the linear equation : 

— + x 0/(p) = //(P) . Solve (a) y = mxp + n (1 + p 3 ) 2, 
dp (p) - p_p_- (p) 

(S) y = x (p + a Vl + p 2 ), (7) x = yp + ap 2 , (5) «/ = (! + p) x + p 2 . 



23^ DIFFERENTIAL EQUATIONS 

7. As any straight line is y = mx + &, any family of lines may be represented as 
y — mx +f(m) or by the Clairaut equation y = px + /(j>). Show that the orthog- 
onal trajectories of any family of lines leads to an equation of the type of Ex. 6. 
The same is true of the trajectories at any constant angle. Express the equations 
of the following systems of lines in the Clairaut form, write the equations of the 
orthogonal trajectories, and integrate : 

(a) tangents to x 2 + y' 2 = 1, (/3) tangents to y 2 = 2 ax, 

(7) tangents to y 2 = x 3 , (8) normals to y 2 = 2 ax, 

(e) normals to y 2 = x 3 , (f) normals to b 2 x 2 + a 2 y 2 = a 2 b 2 . 

8. The evolute of a given curve is the locus of the center of curvature of the 
curve, or, what amounts to the same thing, it is the envelope of the normals of the 
given curve. If the Clairaut equation of the normals is known, the evolute may be 
obtained as its singular solution. Thus find the evolutes of 

(a) y 2 = 4 ax, (£) 2 xy = a 2 , (7) X3 -f yz = aJ, 

(«> 1 + ^=1, (^ = ~. <©>«««■ + «-). 

9. The involutes of a given curve are the curves which cut the tangents of the 
given curve orthogonally, or, what amounts to the same thing, they are the curves 
which have the given curve as the locus of their centers of curvature. Find the 
involutes of 

(a) x 2 + y 2 = a 2 , (/3) y 2 = 2 mx, (7) y = a cosh (x/a). 

10. As any curve is the envelope of its tangents, it follows that when the curve 
is described by a property of its tangents the curve may be regarded as the singu- 
lar solution of the Clairaut equation of its tangent lines. Determine thus what 
curves have these properties : 

(a) length of the tangent intercepted between the axes is /, 

(/3) sum of the intercepts of the tangent on the axes is c, 

(7) area between the tangent and axes is the constant fc 2 , 

( 5 ) product of perpendiculars from two fixed points to tangent is A: 2 , 

( e ) product of ordinates from two points of x-axis to tangent is k 2 . 

dF / 

11. From the relation — = /x \M 2 + N 2 of Proposition 3, p. 212, show that as 

dn 

the curve F = C is moving tangentially to itself along its envelope, the singular 
solution of Mdx + Ndy = may be expected to be found in the equation 1/fi = ; 
also the infinite solutions. Discuss the equation l//x = in the following cases : 



(a) Vl - y 2 dx = Vl - x 2 dy, (/3) xdx + ijdy = Vx 2 + y 2 - a 2 dy. 

102. Equations of higher order. In the treatment of special prob- 
lems (§ 82) it was seen that the substitutions 

dy dhi dp d 2 !/ dp AN 

rendered the differential equations integrable by reducing them to in- 
tegrable equations of the first order. These substitutions or others like 
them are useful in treating certain cases of the differential equation 



ADDITIONAL OEDINARY TYPES 235 

^f(x, y, y', y", •••, y {n) ) = of the nth order, namely, when one of the 
variables and perhaps some of the derivatives of lowest order do not 
occur in the equation. 

/ dhj d i+1 y dy\ A 

Incase ^ B> J, _!,... ,_lj = o, (11) 

y and the first i — 1 derivatives being absent, substitute 

The original equation is therefore replaced by one of lower order. If 
the integral of this be F(x, q) = 0, which will of course contain n — i 
arbitrary constants, the solution for q gives 



q =f(x) and y= J ... J f(x) (dx)\ 



(12) 



The solution has therefore been accomplished. If it were more con- 
venient to solve F(x, q) = for x = <f>(q), the integration would be 



J ■ ■j<i{d*) i = f ■ ■ fs w(g)<ky 



(12') 



and this equation with x = <f> (q) would give a parametric expression 
for the integral of the differential equation. 

x being absent, substitute p and regard p as a function of y. Then 
dy _ d 2 y _ dp d z y d / dp 



dx 1' dx' P dy' dx" P dy\ P dy, 
and ^, P ,^,-..,^ = 0. 

In this way the order of the equation is lowered by unity. If this equa- 
tion can be integrated as F(y, p) = 0, the last step in the solution may 
be obtained either directly or parametrically as 



C dy_ 

J m 



p=m> 77^=* a 4 ) 



It is no particular simplification in this case to have some of the lower 
derivatives of y absent from # = 0, because in general the lower deriva- 
tives of p will none the less be introduced by the substitution that 
is made. 



236 DIEFEKENTIAL EQUATIONS 

' / d*y d 2 y\ 2 /d s y\ 2 „ 
As an example consider [x-— — -— ) =(-— + 1, 
\ dx 3 dx 2 / \dx 3 / 

Then g = X ^ ± \(^) 2+1 and ff = ^i*±Va?+l; 

for the equation is a Clairaut type. Hence, finally, 

y =ff[c 1 x ± Vc 1 2 + l](dx) 2 = i C x x 3 ± J a* Vc 2 + 1 + C 2 x + C 3 . 
As another example consider ?/" — y' 2 = y 2 log y. This becomes 
p < *P-p2 = y2logy or ^5_2i) 2 = 2?/ 2 loo? / . 
The equation is linear in p 2 and has the integrating factor e~ 2 y. 

_p2 e -22/— jy^e- 2 y\ogydy, — p = \e 2, J \ y 2 e~ 2 v\ogydy V , 

and r ^ r = V2 x. 



I \e 2 v Cy 2 e- 2 v\ogydy\ 



The integration is therefore reduced to quadratures and becomes a problem in 
ordinary integration. 

If an equation is homogeneous with respect to y and its derivatives, 
that is, if the equation is multiplied by a power of k when y is replaced 
by ky, the order of the equation may be lowered by the substitution 
y = e z and by taking z' as the new variable. If the equation is homo- 
geneous with respect to x and dx, that is, if the equation is multiplied 
by a power of k when x is replaced by kx, the order of the equation 
may be reduced by the substitution x = e f . The work may be simplified 
(Ex. 9, p. 152) by the use of 

D«y = e~ "<D t (D t - 1) • • . (D t - n + t) y. (15) 

If the equation is homogeneous with respect to x and y and the dif- 
ferentials dx, dy, d 2 y, • • • , the order may be lowered by the substitution 
x = e f , y — e f z, where it may be recalled that 

D«y = e~ «D t (D t -l)...(D t -n + l)y 

= e-<»-v\D t + 1) A • • • (A - n + 2)*. 
Finally, if the equation is homogeneous with respect to x considered of 
dimensions 1, and y considered of dimensions m, that is, if the equation 
is multiplied by a power of k when kx replaces x and k m y replaces y, 
the substitution x = e*, y = e mt z will lower the degree of the equation. 
It may be recalled that 

D£y = e (m - n)t (D t + m) (D t + m — 1) ■ • • (D t -f m — n + l)z. (15") 



ADDITIONAL ORDINARY TYPES 237 



Consider xyy" — xy' 2 = yy' + bxy' 2 /va 2 — x 2 . If in this equation y be replaced 
by ky so that y' and y" are also replaced by ~ky' and ky" ', it appears that the 
equation is merely multiplied by k 2 and is therefore homogeneous of the first 
sort mentioned. Substitute 

y - e s , y' = e 2 z', , y" = e z {z" + z' 2 ). 
Then e 2z will cancel from the whole equation, leaving merely 

xdz' 1 , bxdx 



xz 



= z' + bxz' 2 /Va 2 — x 2 or — dx 



z' z Va 2 - x 2 

The equation in the first form is Bernoulli ; in the second form, exact. Then 

- = b Va 2 - x 2 + C and dz 



6Va 2 -x 2 + C 

The variables are separated for the last integration which will determine z = log y 
as a function of x. 

d 2 u dy 

Again consider x 4 — - = (x 3 + 2xy) 4y 2 . If x be replaced by kx and y by 

dx 2 dx 

k 2 y so that y' is replaced by ky' and y" remains unchanged, the equation is multi- 
plied by k* and hence comes under the fourth type mentioned above. Substitute 

x = e', y = e 2 % D x y = e*(B t + 2) z, D 2 y = (D t + 2) (D t + 1) 2. 

Then e 4t will cancel and leave z" + 2 (1 — z) z' = 0, if accents denote differentiation 
with respect to t. This equation lacks the independent variable t and is reduced 
by the substitution z" = z'dz' '/dz. Then 

^ +2 (l_z) = 0, 2 ' = ^ = (l_ z )2 + c, =dt. 

dz V ; ' dt V ' ' (1 - z 2 ) + C 

There remains only to perform the quadrature and replace z and t by x and ?/. 
103. If the equation may be obtained by differentiation, as 
dy d n y\ cZO dCl cQ , cQ 



it is called an exact equation, and £l(x, y, y', • • -, y^- 1 ^ = C is an inte- 
gral of ^ = 0. Thus in case the equation is exact, the order may be 
lowered by unity. It may be noted that unless the degree of the nth. 
derivative is 1 the equation cannot be exact. Consider 

*(*,y,y',---,y 0,) ) = ^ 0,) + ^, 

where the coefficient of y {rC) is collected into <£ r Now integrate <j> 1} par- 
tially regarding only y^ n ~ ls > as variable so that 



/ 



d cQ, dQ 



f/cc x dx dy { 

That is, the expression ^ — O/ does not contain y (n) and may contain 
no derivative of order higher than n — k, and may be collected as 



238 DIFFERENTIAL EQUATIONS 

indicated. Now if \I> was an exact derivative, so must ^ — Q\ be. Hence 
if m =£ 1, the conclusion is that ^ was not exact. If m = 1, the process 
of integration nia}^ be continued to obtain 2 by integrating partially 
with respect to y( n ~ k - l \ And so on until it is shown that ^ is not exact 
or until & is seen to be the derivative of an expression Q 1 -\- Q 2 + ■ ■ • = C. 

As an example consider & = x 2 y'" + xy" + (2 xy — 1) y' + y 2 = 0. Then 

fy = CxHy" = x 2 y", <& — &[=— xy" + (2 xy — 1) y' + y 2 , 

fi 2 = f— xd V' = — x y', *" — °i;— fi 2 = 2 W' + 2/ 2 = (z?/ 2 )'- 
As the expression of the first order is an exact derivative, the result is 

*■ — o(— fig — ( a; 2/ 2 ) / = ° ; and *"i = x2 y" — x v' + xy 2 — c r = o 

is the new equation. The method may be tried again. 

Oj = f x-%' = xy, ^ - fii = -3 xi/' + xy 2 - C x . 

This is not an exact derivative and the equation < Sr 1 = is not exact. Moreover 
the equation ^ 1 = contains both x and y and is not homogeneous of any type 
except when C\ — 0. It therefore appears as though the further integration of the 
equation SI> = were impossible. 

The method is applied with especial ease to the case of 

^g + A- 1 |a + - + A-„- 1 | / + A-»,-«(x)=0, (17) 

where the coefficients are functions of x alone. This is known as the 
linear equation, the integration of which has been treated only when 
the order is 1 or when the coefficients are constants. The application 
of successive integration by parts gives 

12i = AV-D, 2 = (X 1 - X > ( »- 2 \ H 3 = (X 2 - X[ + X 'V-- 8 >, • • • ; 

and after n such integrations there is left merely 

(A'„ - J,'_i + ' ■ ■ + (- 1)"" 1 ^ + (- 1)»-Y ) y-E, 

which is a derivative only when it is a function of x. Hence 

x n - xu + ■••+(- ly-^ + (- iyx = o (is) 

is the condition that the linear equation shall be exact, and 

Xy-i) + ( Xl - Xi) y^ + (A 2 - X[ + X?)y&-V + • • ■ =JrcIx (19) 
is the first solution in case it is exact. 

As an example take y'" + y" cos x — 2 y' sin x — y cos x = sin 2 x. The test 
X 3 — X 2 ' + X[' — .Xj" = — cosx + 2 .cosx — cosx = 



ADDITIONAL OKDIXARY TYPES 239 

is satisfied. The integral is therefore y" + y'cosx — ysinx ==— \ cos2x + C v 
This equation still satisfies the test for exactness. Hence it may be integrated 
again with the result y' + y cos z = — \ sin 2 x + C 1 x + C 2 . This belongs to the 
linear type. The final result is therefore 

y = e-smx r eS mx(c iX + C 2 ) dx + C 3 e~ sinx + i (1 — sinx). 

EXERCISES 

1. Integrate these equations or at least reduce them to quadratures : 
(a) 2 xy'"y" = y" 2 - a 2 , (/3) (1 + x 2 ) y" + 1 + y' 2 = 0, 

(7) y iv + «V =0, (5) r - mV" = e«*, (e) xhj™ + ft 2 ?/" = 0, 

(f) «W = z, (v) vy" + 2/1=. 0, (0) y"'y" = 4, 

(0 (1 - x 2 )y" - •*./ = 2 > («) ?/ iv = ^V", (X) 2/" =/(y), 

(/*) 2 (2 a - y) y" = 1 + y' 2 , (*>) yy" - if 1 - yhj' = 0, 
(0) yy" + y' 2 + 1 = 0, (w) 2 y" = e^, (p) yV = a. 

2. Carry the integration as far as possible in these cases : 
(a) x 2 y" — (mx 2 y' 2 + ra/ 2 )2, (/3) mx z y" = (y — x?/') 2 , 

( 7 ) &V' =.(y - Z2/') 3 , (5) a- 4 ?/" - xV - xV 2 \±y 2 - 0, 

(e) x- V + x-iy = \y' 2 , (f) ayy" + &2/' 2 = 2/2/'(c 2 + x 2 )~i 

3. Carry the integration as far as possible in these cases : 

(a) (?/ + x)y"' + 6yy'y" + y" + 2 y' 2 = 0, (£) 2/V - ?/xV = xy 2 , 
(7) xW + 3xW + 9xW + 9x 2 ?/ 2 + l$xyy' + Sy 2 = 0, 
(5) 2/ + 3xy' + 2^ ,s + (x 2 + 2y 2 y')y" = 0, 
( e ) (2 x V + x 2 y) y" + 4 x 2 ?/ 2 + 2 xyy' = 0. 

4. Treat these linear equations: 

(a) xy" + 2y = 2x, ((3) (x 2 - l)y" + ±xy' + 2y = 2 x, 

(7) V" — y' cot x + y csc 2 x = cosx, (5) (x 2 — x) y" + (3 x — 2) y' + y = 0, 

(e) (x - x*)y'" + (1 - 5x 2 )y" - 2x?/ + 2y = 6x, 

(t) (X s + ^ 2 -3x+ l)y'" + (9x 2 + 6x- 9)^ + (18 x + 6)2/' + 62/ = x 3 , 

(t?) (jc + 2) 2 ?/ w + (x + 2) y" + y' - 1, (^) iV' + 3 x?/' + ?/ = x, 

(0 (x 3 -x)2/ / ' + (8x 2 -3)?/ / + 14x?/ / +42/ = 0. 

5. Kote that Ex. 4 (#) comes under the third homogeneous type, and that Ex. 4 
(17) may be brought under that type by multiplying by (x + 2). Test sundry of Exs. 
1, 2, 3 for exactness. Show that any linear equation in which the coefficients are 
polynomials of degree less than the order of the derivatives of which they are the 
coefficients, is surely exact. 

6. Sometimes, when the condition that an equation be exact is not satisfied, it 
is possible to find an integrating factor for the equation so that after multiplication 
by the factor the equation becomes exact. Eor linear equations try x m . Integrate 

{a) xhj" + (2 x 4 - x) y' - (2 x 3 - 1) y = 0, (/3) (x 2 - x 4 ) y" - xhj' - 2 y = 0. 

7. Show that the equation y" + Ty' + Qy' 2 = may be reduced to quadratures 
1° when P and Q are both functions of ?/, or 2° when both are functions of x, or 3° 
when P is a function of x and Q is a function of y (integrating factor 1/y'). In 
each case find the general expression for y in terms of quadratures.* Integrate 
y" + 2?/ / cotx + 2y' 2 tdiny = 0. 



240 DIFFERENTIAL EQUATIONS 

8. Find and discuss the curves for which the radius of curvature is proportional 
to the radius r of the curve. 

9. If the radius of curvature R is expressed as a function R = R(s) of the arc s 
measured from some point, the equation R = R (s) or s = s (R) is called the intrinsic 
equation of the curve. To find the relation between x and y the second equation 
may be differentiated as ds = s'(R) dR, and this equation of the third order may be 
solved. Show that if the origin be taken on the curve at the point s = and if the 
x-axis be tangent to the curve, the equations 

r s T r s dsl. r s . I" f s ds~l 

x = I cos | — Ids, y = I sin — \ds 

Jo Uo RJ Jo Uo RJ 

express the curve parametrically. Find the curves whose intrinsic equations are 

(a) R = a, ((3) aR = s 2 + a 2 , (7) R 2 + s 2 = 16 a 2 . 

10. Given F = y(») + X^-D + X 2 i/(»~ 2 ) + • • • + X n ^y' + X n y = 0. Show 
that if /x, a function of x alone, is an integrating factor of the equation, then 

$ = moo - (x l/t )o»-i) + (x aM )<— 2 > + (- iy-\x n _ lfJ .y + (- \)nx ni , = 

is the equation satisfied by /*. Collect the coefficient of /x to show that the condition 
that the given equation be exact is the condition that this coefficient vanish. The 
equation $ — is called the adjoint of the given equation F = 0. Any integral fi 
of the adjoint equation is an integrating factor of the original equation. Moreover 
note that 

f/xFdx = fiyin-V + (txX x - ix')y^-2) +... + (_ l)nj y <$>dx, 

or d[{xF - (- l) n y&] = d [fxy( n -« + ( f xX 1 - p') y(» - 2) + . . .] = dQ. 

Hence if /xF is an exact differential, so is y$. In other words, any solution y of the 
original equation is an integrating factor for the adjoint equation. 

104. Linear differential equations. The equations 

x <P*y + X i° n ~ 1 y + ---+ x n -iDy + X n y =R(x), 
X D"y + Xfi^hj + • • • + X n _ r Dy + X n y = 

are linear differential equations of the nth. order ; the first is called the 
complete equation and the second the reduced equation. If y 1 , y 2 , y z , 
are any solutions of the reduced equation, and C 1? C 2 , C g , • • • are any 

constants, then y = C 1 y 1 + C 2 y 2 + C 3> ?/ 3 H is also a solution of the 

reduced equation. This follows at once from the linearity of the reduced 
equation and is proved by direct substitution. Furthermore if I is any 
solution of the complete equation, then y -f- J is also a solution of the 
complete equation (cf. § 96). 

As the equations (20) are of the nth. order, they will determine y (n) 
and, by differentiation, all higher derivatives in terms of the values of 
x ^y^y\ • • • ? V {n -1) - Hence if the values of the n quantities y , y' Q , • • • , y^ 1 _1) 
which correspond to the value x = x be given, all the higher derivatives 
are determined (§§ 87-88). Hence there are n and no more than n arbi- 
trary conditions that may be imposed as initial conditions. A solution 



ADDITIONAL ORDINARY TYPES 241 

of the equations (20) which, contains n distinct arbitrary constants is 
called the general solution. By distinct is meant that the constants 
can actually be determined to suit the n initial conditions. 
If y v y 2 , • • • , y n are n solutions of the reduced equation, and 

V = C 1 y 1 +C 2 y 2 +... + C n y n , 

y' = Crfi + C 2 y' 2 H f- C n y n , ^l) 

yO.-D = C iy <?-v + C. 2 yi n -» + • • • + C n y^\ 

then y is a solution and y', • • • , y {n ~ ls > are its first n — 1 derivatives. If 
a? be substituted on the right and the assumed corresponding initial 
values y , y' Q , ■ • • , y ( n _1) be substituted on the left, the above n equations 
become linear equations in the n unknowns C v C 2 , • • • , C n ; and if they 
are to be soluble for the C's, the condition 

y 1 y 2 ■■• Vn 
yi yi ■■■ y'n 



w (y v y 2 , ••■,y n ) 



y£-l) yCn-1, ... yj.-l, 



^ (22) 



must hold for every value of x = x Q . Conversely if the condition does 
hold, the equations will be soluble for the C's. 

The determinant W(y 1 , y 2 , • • • , y n ) is called the Wronskian of the n, 
functions y v y 2 , •••, y n . The result may be stated as: If n functions 
y v y 2 , • • • , y n which are solutions of the reduced equation, and of which 
the Wronskian does not vanish, can be found, the general solution of the 
reduced equation can be written down. In general no solution of the 
equation can be found, whether by a definite process or by inspection ; 
but in the rare instances in which the n solutions can be seen by inspec- 
tion the problem of the solution of the reduced equation is completed. 
Frequently one solution may be found by inspection, and it is therefore 
important to see how much this contributes toward effecting the solution. 

If y l is a solution of the reduced equation, make the substitution 
y = y^. The derivatives of y may be obtained by Leibniz's Theorem 
(§8). As the formula is linear in the derivatives of z, it follows that 
the result of the substitution will leave the equation linear in the new 
variable z. Moreover, to collect the coefficient of z itself, it is necessary 
to take only the first term y^z in the expansions for the derivative y a \ 

Hence (A- oi/ <»> + x&t -« + ■•• + a-. _ iy ; + X n!/l ) , = 

is the coefficient of z and vanishes by the assumption that y x is a solu- 
tion of the reduced equation. Then the equation for z is 

V° + p / n _1) + • • • + p n-**" + P-i*' = ; (23) 



242 DIFFERENTIAL EQUATIONS 

and if »' be taken as the variable, the equation is of the order n — 1. 
It therefore appears that the knowledge of a solution y 1 reduces the order 
of the equation by one. 

Now if y 2 , y 3 , ■ • , y p were other solutions, the derived ratios 

s)' -•=©' ■ •-=©' « 

would be solutions of the equation in s' ; for by substitution, 

y = v x z x = Vv y = Ui z 2 = y^ -;> v = 1/^-1 = y P 

are all solutions of the equation in y. Moreover, if there were a linear 
relation C^ -f C 2 z 2 + • • • + C p _ x z p _ x = connecting the solutions «J, 
an integration would give a linear relation 

c> 2 + c ,y, +"-;•■•> c,-i% + c^ = o 

connecting the p solutions T ^. Hence if there is no linear relation (of 
which the coefficients are not all zero) connecting the p solutions y i of 
the original equation, there can be none connecting the p — 1 solutions 
z{ of the transformed equation. Hence a knowledge of p solutions of 
the original reduced equation gives a new reduced equation of which 
p — 1 solutions are known. And the process of substitution may be 
continued to reduce the order further until the order n — p is reached. 

As an example consider the equation of the third order 

(1 - x) y'" + {x 2 - 1) y" - x*y + xy = 0. 
Here a simple trial shows that x and e x are two solutions. Substitute 

y = e x z, y' = e x (z + z% y" = e x (2 z' + «"), y'" = e x (S z" + z"'). 
Then (1 - x)z'" + (x 2 - 3x + 2)z" + (x 2 - 3x + l)z' = 

is of the second order in z' . A known solution is the derived ratio (x/eP°)'. 

z' — (xe~ x y = e~ x (l — x). Let z' = e~ x (l — x)w. 
Prom this, z" and z'" may be found and the equation takes the form 

(1 — x) vo" + (1 -f x) (x — 2) w' = or = xdx dx • 

w' x — 1 

This is a linear equation of the first order and may be solved. 

logiw' = Ix 1 — 21og(x — 1) + C or w' = C 1 e^{x — 1)~ 2 . 

Hence w = C x fei x \x - l)~ 2 dx + C 2 , 

«=o>/(£)'/«**<«-i)-w+ <>.£ + C 3, 

2/ = e*z = c i ex f(-}jf^ ""~( x ~ I)" 2 (^) 2 + CU + C s e«. 



ADDITIONAL OEDIXARY TYPES 243 

The value for y is thus obtained in terms of quadratures. It may be shown that in 
case the equation is of the nth degree with p known solutions, the final result will 
call for p (n — p) quadratures. 

105. If the general solution y = C\y 1 + C 2 y 2 -\ h C n y n of the reduced 

equation has been found (called the complementary function for the 
complete equation), the general solution of the complete equation may 
always be obtained in terms of quadratures by the important and far- 
reaching method of the variation of constants due to Lagrange. The 
question is : Cannot functions of x be found so that the expression 

V = C t (x) y x + C 2 (x) y 2 + --- + C n (x) y n (24) 

shall be the solution of the complete equation ? As there are n of these 
functions to be determined, it should be possible to impose n — 1 condi- 
tions upon them and still find the functions. 

Differentiate y on the supposition that the C's are variable. 

!/' = CiV'i + C 2 y 2 + • • • + C n y n + yi C{.+ y 2 C 2 + • • • + y n C' n . 
As one of the conditions on the C's suppose that 

VxC'x + y^ 2 + ••• + y n C' n = 0. 
Differentiate again and impose the new condition 

y'iC{ + y 2 C 2 + ... + y' n C n =0, 
so that y" = C x y'i + C 2 y 2 + ■ • ■ + C n y';. 

The differentiation may be continued to the (n — l)st condition 

y{ n - 2) c[ + yi n ~ 2) c 2 h h y^'^c'n = 0, 

and y° l _1) = C^y? ~ 1} -f- C 2 yf 2 n ~ 1} -\ \- C n y^ " 1 \ 

Then y™ = C\y[^ + C 2 y^ +■■■ + C n y™ 

+ yi~ l)c 'i + i/^ _1) C'.; H h y^- l) C' n . 

Xow if the expressions thus found for y, y', y", • ••, y (n ~ 1} , y (n) be 
substituted in the complete equation, and it be remembered that y , 
y 2 , • ■ • , y n are solutions of the reduced equation and hence give when 
substituted in the left-hand side of the equation, the result is 

yf-Dcj + y?~ 1} c 2 + • • • + ylr^c: = r. 

Hence, in all, there are n linear equations 



(25) 



yiC[ 


+ y-ic: 2 


+ ■ 


'■ + y n c'n 


= 0, 


y[c[ 


+ y*e 2 


+ •• 


■+y'nC' n 


= 0, 



yi n - 2) c{ + yf-^c 2 + •"■■ + 2/?- 2) c; = o, 



244 DIFFERENTIAL EQUATIONS 

connecting the derivatives of the Cs ; and these may actually be solved 
for those derivatives which will then be expressed in terms of x. The 
Cs may then be found by quadrature. 

As an example consider the equation with constant coefficients 

(D 3 + D) y = sec x with y = C 1 + C 2 cos x + C 3 sin x 

as the solution of the reduced equation. Here the solutions y x ,y 2 , y 8 may be taken 
as 1, cosx, sinx respectively. The conditions on the derivatives of the C's become 
by direct substitution in (25) 

C[ + cosxC 2 + sinxCg = 0, — sinxC 2 + cosxCg = 0, — cosxC^ — sinxCg =secx. 

Hence C[ = sec x, C' % ' = — 1, C' 3 = — tan x 

and C l = log tan (i x + \ it) + c x , C 2 = — x -f c 2 , C 3 = log cos x + c 3 . 

Hence y = c t + log tan (i x + \ tt) + (c 2 — x) cos x + (c 3 + log cos x) sin x 

is the general solution of the complete equation. This result could not be obtained 
by any of the real short methods of §§ 96-97. It could be obtained by the general 
method of § 95, but with little jf any advantage over the method of variation of 
constants here given. The present method is equally available for equations with 
variable coefficients. 

106. Linear equations of the second order are especially frequent in 
practical problems. In a number of cases the solution may be found. 
Thus 1° when the coefficients are constant or may be made constant by 
a change of variable as in Ex. 7, p. 222, the general solution of the 
reduced equation may be written down at once. The solution of the 
complete equation may then be found by obtaining a particular integral 
I by the methods of §§ 95-97 or by the application of the method of 
variation of constants. And 2° when the equation is exact, the solution 
may be had by integrating the linear equation (19) of § 103 of the first 
order by the ordinary methods. And 3° when one solution of the re- 
duced equation is known (§ 104), the reduced equation may be com- 
pletely solved and the complete equation may then be solved by the 
method of variation of constants, or the complete equation may be 
solved directly by Ex. 6 below. 

Otherwise, write the differential equation in the form 

The substitution y = uz gives the new equation 

d 2 z /2du , \dz 1 . „ ' , R ,___ 

yi+-7- + P^ + - (u" + Pu' + Qu) z = — (26') 

dx z \u dx J dx u u 

If a be determined so that the coefficient of z' vanishes, then 

» = *"*/- and g + ( Q -if-ipA, = ^/- (27) 



ADDITIONAL OKDINAKY TYPES 245 

Now 4° if Q — i P' — I P 2 is constant, the new reduced equation in 
(27) may be integrated ; and 5° if it is k/x 2 , the equation may also be 
integrated by the method of Ex. 7, p. 222. The integral of the com- 
plete equation may then be found. (In other cases this method may 
be useful in that the equation is reduced to a simpler form where solu- 
tions of the reduced equation are more evident.) 

Again, suppose that the independent variable is changed to z. Then 

tfy z" + Pz' dy Q„_R , 2 QN 

Now 6° if z' 2 = ±Q will make z" -f Pz' = kz' 2 , so that the coefficient 
of dy/dz becomes a constant k, the equation is integrable. (Trying if 
z' 2 = -j- Qz 2 will make z" -f- Pz' — kz' 2 /z is needless because nothing in 
addition to 6° is thereby obtained. It may happen that if z be deter- 
mined so as to make z" -\- Pz' = 0, the equation will be so far simpli- 
fied that a solution of the reduced equation becomes evident.) 

d 2 v 2 dv a 2 

Consider the example 1 -\ y = 0. Here no solution is apparent. 

dx 2 x dx x 4 

Hence compute Q — \P' — \ P 2 . This is a 2 /x* and is neither constant nor propor- 
tional to 1/x 2 . Hence the methods 4° and 5° will not work. From z' 2 — Q = a 2 /x i 
or z' = a/x 2 , it appears that z" + Pz' = 0, and 6° works ; the new equation is 

^l + y = with z = --. 
dz 2 x 

The solution is therefore seen immediately to be 

y = C 1 cos z — C 2 sin z or y = C x cos (a/x) + C 2 sin (a/x). 

If there had been a right-hand member in the original equation, the solution could 
have been found by the method of variation of constants, or by some of the short 
methods for finding a particular solution if B had been of the proper form. 

EXERCISES 

1. If a relation C x y x + C 2 y 2 + • • • + C n y n = 0, with constant coefficients not all 0, 
exists between n functions y x , y 2 , - ■ • , y n oi x for all values of x, the functions are 
by definition said to be linearly dependent; if no such relation exists, they are said 
to be linearly independent. Show that the nonvanishing of the Wronskian is a 
criterion for linear independence. 

2. If the general solution y = C l y 1 + C 2 y 2 + • • • + C n y n is the same for 
X yW + JT 1 2/(«-i) + --- + X n2 / = and P y<»> + P,y<»-i> + • • • + P n y = 0, 

two linear equations of the nth order, show that y satisfies the equation 
(A\P - X^P^n-D +-... + (X n P - X P n )y = 

of the (n — l)st order ; and hence infer, from the fact that y contains n arbitrary 
constants corresponding to n arbitrary initial conditions, the important theorem : 
If two linear equations of the nth order have the same general solution, the corre- 
sponding coefficients are proportional. 



2-16 * DIFFERENTIAL EQUATIONS 

3. If y x , y 2 , • • • , y n are n independent solutions of an equation of the nth order, 
show that the equation may be taken in the form W(y 1 , y 2 , • • • , y n , y) = 0. 

4. Show that if, in any reduced equation, X n -i + xX n = identically, then x 
is a solution. Find the condition that x m be a solution ; also that e nvc be a solution. 

5. Find by inspection one or more independent solutions and integrate : 
(a) (l + x 2 )y"-2xy'+2 y = 0, ((3) xy" + (I- x)y' - y = 0, 

(7) {ax - bx 2 )y" - ay' + 2by = 0, (8) \y" + xy' - (x + 2)y = 0, 

( e) (log a: + — - — + -V" + (log a; + — + — - -\y" + ( - - -) (y' - xy) = 0, 
\ X 4 X 2 X] \ X 4 X 3 x 2 / \x 2 X) 

(T) 2/ iv -x?/ /// +X2/ / -y = 0, (17) (4x 2 -x + l)?/ ,/ +8xV / -4x?/ / -8?/ = 0. 

6. If ?/ 1 is a known solution of the equation y" + Ty' -{- Qy = R of the second 
order, show that the general solution may be written as 

V = C lVl + C , f. L /*** + Vl Ch e S™* f Vl ef™B(dx)*. 

J y 1 J y x J 

7. Integrate: 

(a) xy"- (2x+ l)y'+ (x + l)y = x 2 — x—% 

(/3) y" - xV + xy = x, (7) xy" + (l — x)y' — y = e*, 

( 5 ) y" — xy' + (x — 1) y = E, ( e ) y" sin 2 x +<y' sin x cos x — y = x — sin x. 

8. After writing down the integral of the reduced equation by inspection, apply 
the method of the variation of constants to these equations : 

(a) (Z> 2 + 1) y = tan x, (/3) (D 2 + 1) y = sec 2 x, (7) (D - lfy = e*(l - x)- 2 , 
(5) (1 - x) y" + xy' - y = (1 - x) 2 , (e) (1- 2x + x 2 )(y'"-l)-x 2 y"+2xy'-y = 1. 

9. Integrate the following equations of the second order: 

(a) 4 x V + 4 xy + (x 2 + l) 2 y = 0, (/3) y" - 2 ?/' tan x - (a 2 + 1) y = 0, 

(7) xy" -\- 2y' — xy = 2e x , (8) y" sin x + 2 y' cos x + 3 y sin x = e*, 

(e) y" + 2/' tan x + ?/ cos 2 x = 0, (£*) (1 — x 2 ) y" — x?/' + 4 y = 0, 

(77) y" + {2 e*~ l)y' + e 2 ^ = e 4 *, (0) x*y" + 3x 5 ?/ + ?/ = x~ 2 . 

10. Show that if X y" + X^' + X 2 y = R may be written in factors as 

' (X Q B 2 + xj> + X 2 ) y = ( Pl D + ffl ) (p 2 D + q 2 )y = B, 

where the factors are not commutative inasmuch as the differentiation in one 
factor is applied to the variable coefficients of the succeeding factor as well as 
to D, then* the solution is obtainable in terms of quadratures. Show that 

QlPo + PlPi + 2>1?2 = X l aild ?1?2 + Pl«2 = X 2 • 

In this manner integrate the following equations, choosing p x andp 2 as factors of 

X and determining g 1 and q 2 by inspection or by assuming them in some form and 

applying the method of undetermined coefficients : 

(a) xy" + (1 - X )y'-y = e* (p) 3 x 2 y" + (2-6 x 2 ) y' - 4 = 0, 

(7) 3xV / +(2 + 6x-6x 2 )?/ / -4?/=:0, (5) (x 2 - l)y"- (3x + 1)2/'- x(x-l)?y = 0, 

(e) axy" + (3a + bx)y' + 3by = 0, (f) xy" - 2x(l + x)?/ + 2(1 + x)?y = x 3 . 

11. Integrate these equations in any manner : 



ADDITIONAL ORDINARY TYPES 247 

(7) V" + y' tan x + y cos 2 x = 0, (5) y"—2in — jy'+ (n 2 — 2 — W=e»» 

( e) (1 - x 2 ) y" - ay' - c 2 y = 0, (f) (a 2 - x 2 ) y" - 8 xy' - 12 y = 0, 

1 /2 , \ ... .. 9-4z . 6-3x 



y = o, 



(v) y " + -ti — y = e* - + iogx-), (0) y y + 

x 2 log x \x 1 3 — x 3 — x 

(0 y" + 2x~h/ - n 2 y = 0, (k) y" - 4xi/ + (4x 2 - 3)y = e* 2 , 

(X) y" + 2 nz/' cot na + (w 2 - n 2 ) y = 0, (/a) y" + 2 (x- 1 + £x~ 2 ) y' + Ax~±y = 0. 

12. If y 1 and y 2 are solutions of y" + Py' + R = 0, show by eliminating Q and 
integrating that r 

y x y-2 - y 2 yi = Ce J 

What if C — ? If C ^ 0, note that y x and y^ cannot vanish together ; and if 
y 1 (a) = y x (b) = 0, use the relation (y 2 yi) a ; (jJ^Vi^b = k>0 to show that as y\ a and 
y[ 6 have opposite signs, y 2 « and y 2 & have opposite signs and hence y 2 (£) = where 
a < £ < &. Hence the theorem : Between any two roots of a solution of an equation 
of the second order there is one root of every solution independent of the given 
solution. What conditions of continuity for y and y' are tacitly assumed here ? 

107. The cylinder functions. Suppose that C n (x) is a function of x 
which is different for different values of n and which satisfies the two 
equations 

€'„_,(*) -C n+1 (x) = 2 £c,(x), C n _ 1 (x)+C n+1 (x) = ^C n (x). (29) 

Such a function is called a cylinder function and the index n is called 
the order of the function and may have any real value. The two equa- 
tions are supposed to hold for all values of n and for all values of x. 
They do not completely determine the functions but from them follow 
the chief rules of operation with the functions. For instance, by addi- 
tion and subtraction, 

£(*) = <?._,(*) - I C„(x) = - x C„(z) - C. +1 (x). (30) 

Other relations which are easily deduced are 

D x [x n C n (ax)~] = ax n C n _ x (ax), D x [x-*CJ(ax)'] = — ax- n C n+1 (x), (31) 

n n — 1 

D x [x 2 C n (-VaxJ\ = \ Vax~ 2 ~C n _ 1 {^/ax\ (32) 

C' (x) = - G x (x), C_ n (x) = (- l) n C n (x), n integral, (33) 

O n (x) K' n (x) - C n (x) K n (x) = C n +1 (x) K n (x) - C n (x) K n +1 (x) = ± , (34) 

where C and K denote any two cylinder functions. 

The proof of these relations is simple, but will be given to show the use of (29). 
In the first case differentiate directly and substitute from (29). 



D x [x n C„(ax)] =x n 
= x n 



aD ax C n (ax) + - C n (ax) 

11 71 

aC n -i(ax) — a— C n (ax) + - C n (ax) . 



ax 



248 DIFFEKENTIAL EQUATIONS 

The second of (31) is proved similarly. For (32), differentiate. 

D x [x 2 C n (Vax)] = -nx~ 2 C n (V^x) + x2 - -v/--D v ^C„(Vax) 

= - Vax 2 -^= C M (Vax) + C w _i(Vax) ^= C„(V«x) • 

^ LVax vax -I 

Next (33) is obtained 1° by substituting for n in both equations (29). 

C_i(x) - C x (x) = 2 Co (x), C_i(x) + C t (x) = 0, hence C ' (x) = - C x (x) ; 

and 2° by substituting successive values for n in the second of (29) written in the 
form xC n -\ + xC n+ i = 2nC n . Then 

xC_i + xC 1 = 0, xC_ 2 + xC = - 2 0_i, xC + xC 2 = 2C 1 , 
xC_ 3 + x(7_i = - 4C_ 2 , xC 1 + xC 3 = 4C 2 , 

xC_4 + xC_ 2 = -6C 3 , xC 2 + xC 4 = 6C s , 

and so on. The first gives C_i = — (7 r Subtract the next two and use (7_i + C t = 0. 
Then C'_2 — C 2 = or C_ 2 = .(— l) 2 ^. Add the next two and use the relations 
already found. Then C_ 3 + C 3 = or C_ 3 = (— 1) 3 C 3 . Subtract the next two, 
and so on. For the last of the relations, a very important one, note first that the 
two expressions become equivalent by virtue of (29) ; for 

11 Ti 

C n K n — C K n = - C n K n — C n K n +i C n K n + C n +\K n • 

x x 

Now — [x(C n+ iK n — C 7 iK n+1 )] = C n+1 K n -*- C n K n+1 + xK n lC n C n+1 ) 

dx \ x ) 

+ xC n+1 r-K n - K n+ A - xK n+1 r- C n - C n+1 \ 
-xC n (K n -^±lK n+1 \ = 0. 

Hence x(C n+ iK n — C n K n +i) = const. = A, and the relation is proved. 

The cylinder functions of a given order n satisfy a linear differential 
equation of the second order. This may be obtained by differentiating 
the first of (29) and combining with (30). 

2 C n = C n _ 1 — C n+1 = - C n _ l — 2 C n H - C n+1 
n 1 

= ~ ( C n-1 + C n+l) ~ ~ ( C n-1 ~ C n+l) ~~ % C n- 

He - e S+sS+^-SV-v *- c -«- < 35 > 

This equation is known as BesseVs equation; the functions C n (x), which 
have been called cylinder functions, are often called BesseVs functions. 
From the equation it follows that any three functions of the same order 
n are connected by a linear relation and there are only two independent 
functions of any given order. 



ADDITIONAL ORDINARY TYPES 249 

By a change of the independent variable, the Bessel equation may 
take on several other forms. The easiest way to find them is to operate 
directly with the relations (31), (32). Thus 

i>»[ar"C,(aj)] = - x~ n C n+1 = -x- ar*- 1 ^, 
T>" x \x n C n (x)J = x n C n+1 -\-x-x n C n + 2 

= - x—'C n+1 + 2'(« + l)ar*r ^ - ar-C,. 

Hence ^ 4 (1 + 2n) ^ 4 y - 
1161106 dx 2 ^ x dx +y ~^ 



d?y { (l-2n) dy 

dx 2 x dx 



Again ^ + ^-^Z^ + 2/= 0, 



y = x-"C,(x). 


(36) 


y = aW„(as). 


(37) 


y = /'C,(2^). 


(38) 



Also xif + (1 + ») y' + y = 0, 

And ay" + (1 - ») y' +■ y = 0, y = a? C M (2 VS). (39) 

In all these differential equations it is well to restrict x to positive values 

n n 

inasmuch as, if n is not specialized, the powers of x, as x n , x~ n . x 2 , x *, are 
not always real. 

108. The fact that n occurs only squared in (35) shows that both 
C n (x) and C_ n (x) are solutions, so that if these functions are inde- 
pendent, the complete solution is y = aC n + bC_ n . In like manner the 
equations (36), (37) form a pair which differ only in the sign of n. 
Hence if H n and H_ n denote particular integrals of the first and second 
respectively, the complete integrals are respectively 

y = aH n + bH_ n x~ 2n and y = aH_ n 4- bH n x 2n ; 

and similarly the respective integrals of (38), (39) are 

y = a l n + bl_ n x~ n and y = al_ n +bl n x n , 

where I n and' J_ n denote particular integrals of these two equations. It 
should be noted that these forms are the complete solutions only when 
the two integrals are independent. Xote that 

I u (x) = x~i n C n {2 V~x\ C n (x) = a *) n In(l **)- (40) 

As it has been seen that C n = (— l) n C_ n when n is integral, it follows 
that in this case the above forms do not give the complete solution. 

A particular solution of (38) may readily be obtained in series by the 
method of undetermined coefficients (§ 88). It is 

I n(x) = X «rf o, = i]{n + 1) ^l ) ... (n + i) > (41) 

as is derived below. It should be noted that I_ n formed by changing 
the sign of n is meaningless when n is an integer, for the reason that 



250 DIFFERENTIAL EQUATIONS 

from a certain point on, the coefficients a t have zeros in the denominator. 
The determination of a series for the second independent solution when 
n is integral will be omitted. The solutions of (35), (36) corresponding 
to I n (x) are, by (40) and (41), 

where the factor n ! has been introduced in the denominator merely to 
conform to usage.* The chief cylinder function C n (x) is J n (x) and it 
always carries the name of Bessel. 



To derive the series for I n (x) write 

I» = %+ a x x + a 2 x 2 + • • • + a k -xx k - x + • • • , 

T n — a t + 2 a 2 x + 3 a 3 x 2 + h (k — 1) a^-ix*- 2 + • • • , 

r;= 2 a 2 + 3 • 2 agx + ■ • • + {k - 1) (k - 2) a A ._ix*-s + 



1 
(1 + n) 

x 



= [a + a x (n + 1)] + x [c^ + a 2 2 ( n + 2 )] + ^ 2 2 + a s 3 ( w + 3 )1 
+•'..-. + x*- l [a k -i + 04fc(n + Jkj] + • •• , 

Hence a + a x (n + 1) = 0, a x + a 2 2 (n + 2) = 0, • ■ • , a*_i + a^ (n + k) = 0, 



or a. 



"2 



n + 1 * 2 (n + 2) 2 ! (n + 1) (n + 2) 

«*= <^>^ 

k\{n + l)---(n + k) 

If now the choice a = 1 is made, the series for I n (x) is as given in (41). 
The famous differential equation of the first order 

xy' — ay + by 2 = ex' 1 , (43) 

known as RiccatVs equation, may be integrated in terms of cylinder functions. 
Note that if n — or c = 0, the variables are separable ; and if b = 0, the equation 
is linear. As these cases are immediately integrable, assume ben ^ 0. By a suitable 
change of variable, the equation takes the form 

d£ 2 \ n/ c2£ n 2 6 d^ 

A comparison of this with (39) shows that the solution is 

a 

v = AI_a(- bet) + Bla (- bc$) • (- bc^ , 

n n 

which in terms of Bessel functions J becomes, by (40), 

v = Z**.[;AJ a (2 V- bct)-+ BJ a (2V-6c^]- 



* If n is not integral, both n\ and (w + i) ! must be replaced (§ 147) byT(w + 1) and 
r(n + i + l). 



ADDITIONAL OEDINAEY TYPES 251 

The value of y may be found by substitution and use of (29). 



. n J a (2x 2 V^~6c/?i) - AJ _ a (2x 2 V- bc/n) 

yj-t* Jt:L -n '" 5 . — m 

J a (2 x 2 V- bc/n) + A J B (2 x 2 V^bc / n) 



where A denotes the one arbitrary constant of integration. 

It is noteworthy that the cylinder functions are sometimes expressible in terms 
of trigonometric functions. For when n = \ the equation (35) has the integrals 

y = A sin x + B cos x and y = xi[ACi(x) + BC_ i(x)]. 

Hence it is permissible to write the relations 

x 2 Ci (x) = sin x, x*C_ 1 (x) = cosx, (45) 

where C is a suitably chosen cylinder function of order \. From these equations 
by application of (29) the cylinder functions of order p + |, where p is any integer, 
may be found. 

Now if Kiccati's equation is such that b and c have opposite signs and a/n is 
of the form p + J, the integral (44) can be expressed in terms of trigonometric 
functions by using the values of the functions C + i just found in place of the J's. 
Moreover if b and c have the same sign, the trigonometric solution will still hold 
formally and may be converted into exponential or hyperbolic form. Thus Riccati's 
equation is integrable in terms of the elementary functions when a/n = p + \ no 
matter what the sign of be is. 

EXERCISES 

1 . Prove the following relations : 

{a) 4C'; = C n -2-2C n +C n + 2, (j8) xC n = 2(n + 1) C n+1 - xC n + 2 , 

(7) 2 3 c;" = C n _ 3 - 3 C n _i + 3 C n + i - C n + 3 , generalize, 

(5) xC n = 2(n + 1) C n+1 - 2(w + 3) C n + S + 2(n + 5) C n + 5 -xC n + 6 . 

2. Study the functions defined by the pair of relations 

P;_i(xj + F n+l (x) = 2-^Fn(x), F n _!(x) - F n+1 (x) = -F n (x) 

especially to find results analogous to (30)-(35). 

3. Use Ex. 12, p. 247, to obtain (34) and the corresponding relation in Ex. 2. 

4. Show that the solution of (38) is y = AI n f (- BI n . 

J x n + L I% 

5. "Write out five terms in the expansions of I , I x , I_ i , e/ , J x . 

[2 1 

6. Show from the expansion (42) that \ ! \ / - J Ax) = - sin x. 

7. From (45), (29) obtain the following : 

x?Cs(x) = cosx, x^C-^(x) = I— — 1) sinx cosx, 

i^ , v cosx i 3 . /3 ' \ 

x2C_ a (x) = — smx , X2C 5 (x) = -smx+ 1 cosx. 

2 x ' 2 x \x 2 / 



252 DIFFERENTIAL EQUATIONS 

8. Prove by integration by parts : ( - 2 - — dx = -^+6-±+6-8 f -^— . 

J x 3 x 3 x 4 J x 5 

9. Suppose C n (x) and iT w (x) so chosen that A = 1 in (34). Show that 

2/ = A C n (x) + £JBT ra (x) + L \ K n(x) f ^- dx - C n (x) f 5^ dxl 

is the integral of the differential equation x 2 y" + xy' + (x 2 — n 2 )y = Lx~ 2 . 

10. Note that the solution of Riccati's equation has the form 

v = mtAG% ' andshowffiat | +*<*)» +.«<»)»■=*<*) 

will be the form of the equation which has such an expression for its integral. 

11. Integrate these equations in terms of cylinder functions and reduce the 
results whenever possible by means of Ex. 7 : 

(a) xy' — 5 y + y 2 + x 2 = 0, (p) xy' — 3 y + ?/ 2 = x 2 , 

(7) 2/" + ye 2x = 0, (5) xV + nxy' + (6 + ex 2 ™) y = 0. 

12. Identify the functions of Ex. 2 with the cylinder functions of ix. 

13. Let (x 2 - 1) P; = (n + 1) (P B +i - xP„), JPJ, +1 = xP' n + (n + 1) P n (46) 

be taken as defining the Legendre functions P n (x) of order n. Prove 

(or) (x 2 - 1) P; = n (xP n - P n _i), 08) (2 n + 1) xP M = ■ (n + 1) P M +i + nP M _i , 
(7) (2n + l)P.=P; + i-P;_i, (5) (l-x 2 )P; / -2xP; + n(n + l)P w = 0. 

14. Show that P„Q„-P»Q„ = -^— and P w Q w+ i - P„+iQ„ = -^- , 

x 2 — 1 n+1 

where P and Q are any two Legendre functions. Express the general solution of 
the differential equation of Ex. 13 (S) analogously to Ex. 4. 

15. Let u = x 2 — 1 and let D denote differentiation by x. Show 

Dn+l u n+l - Dn+l(uun) = ul> + 1 U n + 2 (ft + 1) xB n VL n + ft (ft + l)D»-lu», 

jya +i u n +1 _ DnD u n +1 = 2 (ft + 1) J>(xu n ) = 2 (ft + 1) xD n u n + 2 ft (ft + 1) D» -*u». 

Hence show that the derivative of the second equation and the eliminant of P"- 1 ^" 
between the two equations give two equations which reduce to (46) if 

_, , N 1 d n , _ „. f When n is integral these are 

P (x) = (X 2 l) n . -j 

2 n -n\dx n \Legendre's polynomials. 

16. Determine the solutions of Ex. 13 (3) in series for the initial conditions 

(a) p n (0) = i, p» = o, 08) p„(0) = o, , p;(0) = 1. 

17. Take P — 1 and P r = x. Show that these are solutions of (46) and compute 
P 2 , P s , P 4 from Ex. 13 (/3). If x = cos 0, show 

P 2 = f cos20 + i, P 3 = f cos3# + fcostf, P 4 = ff cos4# + ff cos2 + ¥ 9 ? . 

18. Write Ex. 13 (5) as — [(1 - x 2 ) P'l + n(n + 1) P n = and show 

dx 

r +i r +i[ d(i-x 2 )p; fZ(i-x 2 )P;i 

[m(m+l)-n(n+l)]j^ P H PJte=J P m dx Pn dT~^\ 



ADDITIONAL ORDINARY TYPES 253 

Integrate by parts, assume the functions and their derivatives are finite, and show 
/ P n P m dx = 0, if n ?£ m. 

19. By successive integration by parts and by reduction formulas show 

«/-i 22»(n!) 2 J-i dx» dx n 2"-n\J-i v 

/. +1 ; 2 

and I P;dx = , n integral. 

J_i " 2n + l 

/, +i /» +i d n (x 2 — l) n 

20. Show f x m P n cZx = / x™— ^ '- = 0, if m<n. 

J-i J-i dx n 

Determine the value of the integral when m = n. Cannot the results of Exs. 18, 19 
for in and n integral be obtained simply from these results ? 

21. Consider (38) and its solution I = 1 — x -1 — — — -\ — . • . when 

n = 0. Assume a solution of the form y = I Q v + miso that 

d 2 w dw , , _ dl dv .. d 2 v dv 

x 1 f-w + 2x — ± — =0, if x 1 = 0, 

dx 2 dx dx dx dx 2 dx 

is the equation for w if v satisfies the equation xv" + x>' = 0. Show 

a -n^ „ c-n 2 B * 2 %& 2 B ^ 

v = A + B loo- x, xw + w -\- vj = 2B 1 f- • • • . 

2! 2!3! 3!4! 

By assuming w = a x x + a 2 x 2 + • • • , determine the a's and hence obtain 

nT ,r x 2 / 1\ x 3 /, 1 1\ x 4 / 1 1 1\ "I 

w = 2B\ x 1 + -M M+- + - 1 + -H h -)+••• ; 

L 2!*\ 2/ 3-! 2 \- 2 3/ 4! 2 \234/ J' 

and (A + B log x) I + w is then the complete solution containing two constants. 
As AI is one solution, .Blogx • I + w is another. From this second solution for 
n = 0. the second solution for any integral value of n may be obtained by differ- 
entiation ; the work, however, is long and the result is somewhat complicated. 



CHAPTER X 

DIFFERENTIAL EQUATIONS IN MORE THAN TWO VARIABLES 

109. Total differential equations. An equation of the form 

P (x, y, z) dx + Q (x, y, z) cly + R (x, y, z) dz = 0, (1) 

involving the differentials of three variables is called a total differen- 
tial equation. A similar equation in any number of variables would 
also be called total; but the discussion here will be restricted to the 
case of three. If definite , values be assigned to x, y, z, say a, b, c, the 
equation becomes 

Adx + Bdy + Cdz = A (x - a) -f- B(y - b) + C (z - e) = 0, (2) 

where x, y, z are supposed to be restricted to values near a, b, c, and 
represents a small portion of a plane passing through (a, b, e). From 
the analogy to the lineal element (§ 85), such a portion of a plane may 
be called a planar element. The differential equation therefore repre- 
sents an infinite number of planar elements, one passing through each 
point of space. 

Now any family of surfaces F(x, y, z) = C also represents an infinity 
of planar elements, namely, the portions of the tangent planes at every 
point of all the surfaces in the neighborhood of their respective points 
of tangency. In fact 

dF = F' x dx + Fydy + F' z dz == (3) 

is an equation similar to (1). If the planar elements represented by 
(1) and (3) are to be the same, the equations cannot differ by more 
than a factor fx(x, y, z). Hence 

F' x = fiP, F; = fiQ, F' z = pR. 

If a function F(x, y, z) = C can be found which satisfies these condi- 
tions, it is said to be the integral of (1), and the factor /x (x, y, z) by 
which the equations (1) and (3) differ is called an integrating factor 
of (1). Compare § 91. 

It -may happen that /x = 1 and that (1) is thus an exact differential. 
In this case the conditions 

254 



MOEE THAN TWO VARIABLES 255 

which arise from F^ = F^., F' y ' z = F' z ' y , F^ = F^ ZJ must be satisfied. 
Moreover if these conditions are satisfied, the equation (1) will be 
an exact equation and the integral is given by 

F(x, y, ») = J P (x, y,z)dx + J Q (x Q , y, z) dy + I R (x Q , y , z) dz = C, 

Jx Jy J 

where x Q , y , z may be chosen so as to render the integration as simple 
as possible. The proof of this is so similar to that given in the case of 
two variables (§ 92) as to be omitted. In many cases which arise in 
practice the equation, though not exact, may be made so by an obvious 
integrating factor. 

As an example take zxdy — yzdx + x 2 dz = 0. Here the conditions (4) are not 
fulfilled but the integrating factor l/x 2 z is suggested. Then 



xdy — ydx dz _ 

; r — t* 



(I+M 



X' z 

is at once perceived to be an exact differential and the integral is y/x + log z = G. 
It appears therefore that in this simple case neither the renewed application of the 
conditions (4) nor the general formula for the integral was necessary. It often 
happens that both the integrating factor and the integral can be recognized at once 
as above. 

If the equation does not suggest an integrating factor, the question 
arises, Is there any integrating factor ? In the case of two variables 
(§ 94) there always was an integrating factor.. In the case of three 
variables there may be none. For 

R, 



F" - 

xy 




dp 


F" 

yx 


CX 


+ At — > 

cx 


F" - 

yz 


cz 


»Tz = 


K 




dR 


K> 


CX 


dR 

/* — = 

CX 


F" 

xz 


cz 


dp 



Q. 

If these equations be multiplied by R, P, Q and added and if the result 
be simplified, the condition 

w_«\ +a /e_wv +Jl /»-«\_ (5) 

cz cyj \cx cz] \cy cx) v J 

is found to be imposed on P, Q, R if there is to be an integrating fac- 
tor. This is called the condition of integrability. For it may be shown 
conversely that if the condition (5) is satisfied, the equation may be 
integrated. 

Suppose an attempt to integrate (1) be made as follows : First assume 
that one of the variables is constant (naturally, that one which will 



256 DIFFERENTIAL EQUATIONS 

make the resulting equation simplest to integrate), say z. Then 
Pdx + Qdy = 0. Now integrate this simplified equation with an inte- 
grating factor or otherwise, and let F(x, y, z) = cj>(z) be the integral, 
where the constant C is taken as a function <f> of z. Next try to deter- 
mine <j> so that the integral F(x, y, z) = ^>{z) will satisfy (1). To do 
this, differentiate ; 

F' x dx + F' y dy + F' z dz = d<f>. 

Compare this equation with (1). Then the equations* 

F x = XP, Fy = XQ, (F; - XR) dz = dcj> 

must hold. The third equation (F' z — XR) dz = d<j> may be integrated 
provided the coefficient 5 = F' z —- XR of dz is a function of z and <f>, 
that is, of z and F alone. This is so in case the condition (5) holds. It 
therefore appears that the integration of the equation (1) for which (5) 
holds reduces to the succession of two integrations of the type discussed 
in Chap. VIII. 

As an example take (2 x 2 + 2 xy + 2 xz 2 + 1) dx + dy + 2 zdz = 0. The condition 

(2x 2 +2xy + 2xz 2 + 1)0 + 1(- 4xz) + 2z(2x) = 

of integrability is satisfied. The greatest simplification will be had by making x 
constant. Then dy + 2 zdz = and y + z 2 = <j> (x) . Compare 

dy + 2 zdz = d<£ and (2 x 2 + 2 xy + 2 xz 2 + 1) dx + dy + 2 ztfe = 0. 
Then \ = 1, - (2 x 2 + 2 xy + 2 xz 2 + 1) dx = d<f> ; 

or — (2 x 2 + 1 + 2 x<f>) dx = d<j> or d<p + 2 x0dx = — (2 x 2 + 1) dx. 

This is the linear type with the integrating factor e* 2 . Then 

e*\d<p + 2 x0dx) = - e* 2 (2 x 2 + 1) dx or e* 2 = - f e*\2 x 2 + 1) dx + C. 
Hence y + z 2 + e~* 2 f e* 2 (2x 2 + 1) dx = Ce~ x2 or e* 2 (y + z 2 ) + /V 2 (2x 2 + 1) dx = C 

is the solution. It may be noted that e 3 ? is the integrating factor for the original 
equation : 

e* 2 [(2 x 2 + 2 xy + 2 xz 2 + 1) dx + dy + 2 zdz] = d e* 2 (?y + z 2 ) + /V 2 (2 x 2 + 1) dx | . 

To complete the proof that the equation (1) is integrable if (5) is satisfied, it is 
necessary to show that when the condition is satisfied the coefficient S = F' z — \E 
is a function of z and F alone. Let it be regarded as a function of x, F, z instead 
of x, y, z. It is necessary to prove that the derivative of S by x when F and z are 
constant is zero. By the formulas for change of variable* 

'8S\ _ (dS\ /dS\ dF_ /dS\ _ /dS\ cF_ 

)y,z \dx/F,z \SF/ dx ' \dy/ x , z \dF/ XtZ dy 

* Here the factor X is not an integrating factor of (1) , but only of the reduced equation 
Pdx + Qdy = 0. 



MORE THAX TWO VARIABLES 257 

But K = \P and F' = \Q, and hence Q (— ) -P (—) =q(—) ■ 

J \cx/, MZ \ey/ x , z \cx/ FyZ 

(—] -JL(—-\r\- — dXR - axp dXR 

\dx/y, z dx\ez ] czdx ex cz ex 

Hence (^ = X l* - ™) + P^ - B £, 

\dx/y t z \cz dxj cz ex 

and m =x(28_«\ + < ,»la» 

ThenQ^ -p(^) = x|W^ - B + P (« - ?SY| - p["<^- P?- X l 
\ax/ y , 2 W/x,* L \cz ex J \dy dz J A L dx dy_\ 

and «H = xr Q (^-a+p(^-i«) +B (^_^l 

\dx/F,z L \as ex/ \8y cz/ \dx cy/J 

_ E raxQ_?xPl 

L £x £?/ J 

where a term has been added in the first bracket and subtracted in the second. 
Now as X is an integrating factor for Pdx + Qdy, it follows that (\QY X = (^P)' y ; and 
only the first bracket remains. By the condition of integrability this, too, vanishes 
and hence S as a function of x, F, z does not contain x but is a function of F and 
z alone, as was to be proved. 

110. It has been seen that if the equation (1) is integrable, there is 
an integrating factor and the condition (5) is satisfied ; also that con- 
versely if the condition is satisfied the equation may be integrated. 
Geometrically this means that the infinity of planar elements defined 
by the equation can be grouped upon a family of surfaces F(x, y, z) = C 
to which they are tangent. If the condition of integrability is not satis- 
fied, the planar elements cannot be thus grouped into surfaces. Never- 
theless if a surface G (x, y, z) = be given, the planar element of (1) 
which passes through any point (x , y Q , z ) of the surface will cut the 
surface G = in a certain lineal element of the surface. Thus upon the 
surface G (jc, y, z) = there will be an infinity of lineal elements, one 
through each point, which satisfy the given equation (1). And these 
elements may be grouped into curves lying upon the surface. If the 
equation (1) is integrable, these curves will of course be the intersections 
of the given surface G = with the surfaces F = C defined by the 
integral of (1). 

The method of obtaining the curves upon G (x, y, z) = which are 
the integrals of (1), in case (5) does not possess an integral of the form 
F(x, y, z) = C, is as follows. Consider the two equations 

Pdx + Qdy + Rdz = 0, G' x dx + G y dy + G' z dz = 0, 

of which the first is the given differential equation and the second is 
the differential equation of the given surface. From these equations 



258 DIFFERENTIAL EQUATIONS 

one of the differentials, say dz, may be eliminated, and the correspond- 
ing variable z may also be eliminated by substituting its value obtained 
by solving G (x, y, z) = 0. Thus there is obtained a differential equa- 
tion Mdx + Ndy = connecting the other two variables x and y. The 
integral of this, F(x, y) = C, consists of a family of cylinders which cut 
the given surface G = in the curves which satisfy (1). 

Consider the equation ydx + xdy — (x + y + z) dz = 0. This does not satisfy the 
condition (5) and hence is not completely integrable ; but a set of integral curves 
may be found on any assigned surface. If the surface be the plane z = x + y, then 

ydx + xdy — (x + y + z) dz = and dz = dx + dy 

give (x + z) dx + (y + z) dy = or (2 x + y) dx + (2 y + x) dy = 

by eliminating dz and z. The resulting equation is exact. Hence 

x 2 + xy + y 2 — C and z = x + y 

give the curves which satisfy the equation and lie in the plane. 

If the equation (1) were integrable, the integral curves may be used to Obtain 
the integral surfaces and thus to accomplish the complete integration of the equa- 
tion by Mayer's method. For suppose that F(x, y,z) = C were the integral surfaces 
and that F(x, y, z) = F(0, 0, z ) were that particular surface cutting the z-axis at z . 
The family of planes y = Xx through the z-axis would cut the surface in a series 
of curves which would be integral curves, and the surface could be regarded as 
generated by these curves as the plane turned about the axis. To reverse these 
considerations let y = Xx and dy = \dx ; by these relations eliminate dy and y from 
(1) and thus obtain the differential equation Mdx + Ndz = of the intersections 
of the planes with the solutions of (1). Integrate the equation as/(x, z, X) = C and 
determine the constant so that/(x, z, X) =/(0, z , X). Tor any value of X this gives 
the intersection of F(x, y, z) = F(0, 0, z ) with y = Xx. Now if X be eliminated by 
the relation X = y/x, the result will be the surface 

f(x]z,^j--f(o,z ,^j, equivalent to F(x, y, z) = >(0, 0, z ), 

which is the integral of (1) and passes through (0, 0, z ). As z is arbitrary, the 
solution contains an arbitrary constant and is the general solution. 

It is clear that instead of using planes through the z-axis, planes through either 
of the other axes might have been used, or indeed planes or cylinders through any 
line parallel to any of the axes. Such modifications are frequently necessary owing 
to the fact that the substitution /(0, z , X) introduces a division by or a log or 
some other impossibility. For instance consider 

y 2 dx + zdy — ydz = 0, y = Xx, dy — \dx, \ 2 x 2 dx + \zdx — \xdz = 0. 

Then \dx -\ = 0, and Xx =/(%, z, X). 

x 2 x 

But here /(0, z , X) is impossible and the-solution is illusory. If the planes (y — 1) = Xx 
passing through a line parallel to the z-axis and containing the point (0, 1, 0) had 
been used, the result would be 

dy = Xdx, (1 + Xx) 2 cZx + \zdx — (1 + Xx) dz = 0, 



MORE THAN TWO VARIABLES 259 

dx H — - — = 0, and x — =/(x, z, X). 



(1 + Xx) 2 

z 


~ z o 


, at 

or 


X - 


s 

y 


1+ Xx 


1 + Xx 


^0 



Hence 

is the solution. The same result could have been obtained with x = \z or y = X (x — a). 
In the latter case, however, care should be taken to use/(x, z, X) =/(a, z , X). 

EXERCISES 

1. Test these equations for exactness ; if exact, integrate ; if not exact, find an 
integrating factor by inspection and integrate : 

(a) (y + z) dx + (z + x) dy + (x + y) dz = 0, (/3) y 2 dx + zdy - ydz = 0, 

(7) xdx + ydy - Va 2 - x 2 - y 2 dz = 0, (5) 2 z {dx - cfy) + (x - ?/) dz = 0, 

( e ) (2 x + if- + 2 xz) dx + 2 x?/di/ + x 2 dz = 0, (f ) zydx = zxdy + 2/ 2 dx, 

( v ) x(y-l){z-l)dx + y{z- 1) (x- l)dy + z(x- 1) (y - l)dz = 0. 

2. Apply the test of integrability and integrate these : 

(a) (x 2 — y 2 — z 2 ) dx + 2 xi/di/ + 2 xzdz = 0, 

(/3) (x + y 2 + z 2 + l)dx + 2ydy + 2zdz = 0, 

(7) (2/ + a) 2 dx + afy = {y + a) dz, 

( 5 ) (1 — x 2 - 2 i/ 2 z) dz = 2 xzdx + 2 yz 2 dy, 

( e ) x 2 dx 2 + y 2 dy 2 - z 2 dz 2 + 2 x?/dxd?y = 0, 

(f ) z (xdx + i/dy + zdz) 2 = (z 2 — x 2 — y 2 ) (xdx + ydy + zdz) dz. 

3. If the equation is homogeneous, the substitution x = uz, ?/ = cz, frequently 
shortens the work. Show that if the given equation satisfies the condition of inte- 
grability, the new equation will satisfy the corresponding condition in the new 
variables and may be rendered exact by an obvious integrating factor. Integrate : 

(a) (y 2 + yz) dx + (xz + z 2 ) dy + (y 2 - xy) dz = 0, 

OS) (x 2 y - y 3 - y 2 z) dx + {xy 2 - x 2 z - x 3 ) dy + {xy 2 + x 2 y) dz = 0, 

(7) {U 2 + yz + z 2 ) dx + (x 2 + xz + z 2 ) dy + (x 2 + xy + ?/ 2 ) dz = 0. 

4. Show that (5) does not hold ; integrate subject to the relation imposed : 
(a) ydx + xdy — (x + y + z) dz = 0, x + y + z = k or y = kx, 

(/3) c (xd# + 2/%) + Vl - a 2 x 2 - b 2 y 2 dz = 0, a 2 x 2 + b 2 y 2 + c 2 z 2 = 1, 
(7) dz — aydx + bdy, y — kx or x 2 + y 2 + z 2 = 1 or y =f(x). 

5. Show that if an equation is integrable, it remains integrable after any change 
of variables from x, y, z to it, u, io. 

6. Apply Mayer's method to sundry of Exs. 2 and 3. 

7. Find the conditions of exactness for an equation in four variables and write 
the formula for the integration. Integrate with or without a factor : 

{a) (2 x + y 2 + 2 xz) dx -f 2 xydy + x 2 dz + du = 0, 

(|8) yzudx + xzwdi/ + xyudz + xyzdw =0, 

(7) {y + z + m) dx + (x + z + u) dy + (x + y + w) dz + (x + ?/ + z) dit = 0, 

(S). u {y + z)dx + u(y + z + l)dy + udz - (y ■+ z)du = 0. 

8. If an equation in four variables is integrable, it must be so when any one of 
the variables is held constant. Hence the four conditions of integrability obtained 
by writing (5) for each set of three coefficients must hold. Show that the conditions 



260 DIFFEBENTIAL EQUATIONS 

are satisfied in the following cases. Find the integrals by a generalization of the 
method in the text by letting one variable be constant and integrating the three 
remaining terms and determining the constant of integration as a function of the 
fourth in such a way as to satisfy the equations. 

{a) z(y + z)dx + z(u — x)dy + y (x — u)dz + y (y + z)du = 0, 
(j8) uyzdx + uzx log xdy + uxy log xdz — xdu = 0. 

9. Try to extend the method of Mayer to such as the above in Ex. 8. 

10. If G(x, y, z) = a and H(x, y, z) = b are two families of surfaces defining a 
family of curves as their intersections, show that the equation 

(P'yK - G' Z H;) dx + (G Z H' X - G' x II' z ) dy + (G X H; - G' y II> x ) dz = 

is the equation of the planar elements perpendicular to the curves at every point 
of the curves. Find the conditions on G and H that there shall be a family of sur- 
faces which cut all these curves orthogonally. Determine whether the curves below 
have orthogonal trajectories (surfaces) ; and if they have, find the surfaces : 

(a) y = x + a, z = x + b, (/3) y = ax + 1, z = &x, 

(7) z 2 + y 2 = a 2 , z = b, (5) xy = a, xz = 6, 

( e) x 2 + y 2 + z 2 = a 2 , xy = &, (f ) x 2 + 2 y 2 + 3 z 2 = a, xy + z = 6, 

(77) log xy — az, x + y + z = 5, (9) y = 2 ax + a 2 , z = 2 bx + b 2 . 

11. Extend the work of proposition 3, § 94, and Ex. 11, p. 234, to find the normal 
derivative of the solution of equation (1) and to show that the singular solution may 
be looked for among the factors of /a -1 = 0. 

12. If F = Pi + Qj + Bk be formed, show that (1) becomes F.dr = 0. Show 
that the condition of exactness is VxF = by expanding VxF as the formal vector 
product of the operator V and the vector F (see § 78). Show further that the condi- 
tion of integrability is F.(VxF) = by similar formal expansion. 

13. In Ex. 10 consider VG and V-ET. Show these vectors are normal to the sur- 
faces G = a, H =b, and hence infer that (VCr)x(ViT) is the direction of the inter- 
section. Finally explain why di»(VGxVH) = is the differential equation of the 
orthogonal family if there be such a family. Show that this vector form of the family 
reduces to the form above given. 

111. Systems of simultaneous equations. The two equations 

■£ = /(*, y, z), c ~f x = v ( x > y> z ) ( 6 ) 

in the two dependent variables y and z and the independent variable x 
constitute a set of simultaneous equations of the first order. It is more 
customary to write these equations in the form 

dx _ dy _ dz ^ 
X(x, ij, z) Y(x, y, z) Z (x, y, z) ' 

which is symmetric in the differentials and where X:Y:Z = l:f:y. 
At any assigned point x , y , z of space the ratios dx:dy:dz of the 
differentials are determined by substitution in (7). Hence the equations 



MOEE THA:N T TWO VARIABLES 261 

fix a definite direction at each point of space, that is, they determine a 
lineal element through each point. The problem of integration is to 
combine these lineal elements into a family of curves F(x, y, z) = C , 
G(x, y, z) = C 2 , depending on two parameters C 1 and C' 2 , one curve pass- 
ing through each point of space and having at that point the direction 
determined by the equations. 

For the formal integration there are several allied methods of pro- 
cedure. In the first place it may happen that two of 
dx _dy dy _ dz dx _ dz 

~x~Y' ~y~~z' ~x~~z 

are of such a form as to contain only the variables whose differentials 
enter. In this case these two may be integrated and the two solutions 
taken together give the family of curves. Or it may happen that one 
and only one of these equations can be integrated. Let it be the first 
and suppose that F(x, y) = C 1 is the integral. By means of this inte- 
gral the variable x may be eliminated from the second of the equations 
or the variable y from the third. In the respective cases there arises 
an equation which may be integrated in the form G (y, z, C ) = C 2 or 
G(x, z, F) = C 2 , and this result taken with F(x,y) = C 1 will determine 
the family of curves. 

scdx ijdu dx 
Consider the example — = — - = — Here the two equations 
yz xz y 

xdx ydy . xdx 

= ^-^- and = dz 

y x z 

are integrable with the results x 3 — y 3 = C x , x 2 — z 2 = C 2 , and these two integrals 
constitute the solution. The solution might, of course, appear in very different 
form ; for there are an indefinite number of pairs of equations F(x, y, z, C\) = 0, 
G (x, y, z, C 2 ) == which will intersect in the curves of intersection of x 3 — y 3 = C\ , 
and x 2 — z 2 — C 2 . In fact (if + C x ) 2 = (z 2 + C 2 ) 2 is clearly a solution and could 
replace either of those found above. 

Consider the example = — — = ■ Here 

x 2 -y 2 -z 2 2xy 2xz 

— — — , with the integral y — C,z, 

y z 

is the only equation the integral of which can be obtained directly. If y be elimi- 
nated by means of this first integral, there results the equation 

*L -*L or 2xzdx + [(C 2 +l)z 2 -x 2 ldz = Q. 

This is homogeneous and may be integrated with a factor to give 
x 2 + (Cl + l)z 2 = C 2 z or x 2 + y 2 + z 2 = C 2 z. 
Hence y = C x z, x 2 + y 2 + z 2 = C 2 z 

is the solution, and represents a certain family of circles. 



262 DIFFERENTIAL EQUATIONS 

Another method of attack is to use composition and division. 

dx _ dy _ dz _ Xdx -f- l^dij + vdz 

T~~T ~~~Z~ XX + pY+vZ O 

Here A, /x, v may be chosen as any functions of (x, y, z). It may be 
possible so to choose them that the last expression, taken with one of 
the first three, gives an equation which may be integrated. With this 
first integral a second may be obtained as before. Or it may be that 
two different choices of A, /x, v can be made so as to give the two desired 
integrals. Or it may be possible so to select two sets of multipliers that 
the equation obtained by setting the two expressions equal may be 
solved for a first integral. Or it may be possible to choose A, fx, v so 
that the denominator XX + fx Y + vZ = 0, and so that the numerator 
(which must vanish if the denominator does) shall give an equation 

Xdx -f- fxdy + vdz == (9) 

which satisfies the condition (5) of integrability and may be integrated 
by the methods of § 109. 

Consider the equations = = Here take X, p, v 

x 2 + y 2 + yz x 2 + y 2 — xz (x + y)z 

as 1, — 1, — 1 ; then XX + fiY + vZ = and dx — dy — dz = is integrable as 

x — y — z = C x . This may be used to obtain another integral. But another choice 

of X, fM, v as x, y, 0, combined with the last expression, gives 

xdx + ydz = dz or g{x 2 + y 2 ) = i ogz 2 + c 

{x 2 + y 2 ){x + y) {x+y)z oV ^ y) * ^ 2 

Hence x — y — z = C x and x 2 + y 2 = C 2 z 2 

will serve as solutions. This is shorter than the method of elimination. 

It will be noted that these equations just solved are homogeneous. The substi- 
tution x = uz, y = vz might be tried. Then 

udz + zdu _ vdz + zdv _ dz zdu zdv 

U 2 + V 2 + V U 2 + V 2 — U U + V V 2 — UV + V u 2 — uv — u 

du dv dz 



u 2 — uv + v u 2 — uv — u z 

Now the first equations do not contain z and may be solved. This always happens 
in the homogeneous case and may be employed if no shorter method suggests itself. 

It need hardly be mentioned that all these methods apply equally to 
the case where there are more than three equations. The geometric 
picture, however, fails, although the geometric language may be contin- 
ued if one wishes to deal with higher dimensions than three. In some 
cases the introduction of a fourth variable, as 

dx _ dy _ dz _ dt _ dt ,. ~ 

~X~~Y~~Z~~\ ° Y ~~t' ^ ' 



MOKE THAN TWO VARIABLES 263 

is useful in solving a set of equations which originally contained only 
three variables. This is particularly true when X, Y, Z are linear with 
constant coefficients, in which case the methods of § 98 may be applied 
with t as independent variable. 

112. Simultaneous differential equations of higher order, as 



L v ^<k\ *m. = y(x v — d A 
\ x ' y 'dtdt)' df Y x> y >dtdt) 



d 2 x _ / dx dy\ d?y _ 



^_m)*=n(r,+,%m, -lium^Jr,^,^ 



df r \dt J \ ' ^' dt dt) r dt\ dt ) \ ' ^ 7 dt dt 

especially those of the second order like these, are of constant occur- 
rence in mechanics ; for the acceleration requires second derivatives 
with respect to the time for its expression, and the forces are expressed 
in terms of the coordinates and velocities. The complete integration of 
such equations requires the expression of the dependent variables as 
functions of the independent variable, generally the time, with a num- 
ber of constants of integration equal to the sum of the orders of the 
equations. Frequently even when the complete integrals cannot be 
found, it is possible to carry out some integrations and replace the 
given system of equations by fewer equations or equations of lower 
order containing some constants of integration. 

No special or general rules will be laid down for the integration of 
systems of higher order. In each case some particular combinations of 
the equations may suggest themselves which will enable an integration 
to be performed.* In problems in mechanics the principles of energy, 
momentum, and moment of momentum frequently suggest combinations 
leading to integrations. Thus if 

x" = X, if = Y, z" = Z, 

where accents denote differentiation with respect to the time, be multi- 
plied by dx, dy, dz and added, the result 

x"dx + y"dy + z"dz = Xdx -f Ydy + Zdz (11) 

contains an exact differential on the left ; then if the expression on the 
right is an exact differential, the integration 

h 0*'' 2 + I/' 2 + z" 1 ) = fxdx + Ydy + Zdz + C (11') 

* It is possible to differentiate the given equations repeatedly and eliminate all the 
dependent variables except one. The resulting differential equation, say in x and t, may 
then be treated by the methods of previous chapters ; but this is rarely successful except 
when the equation is linear. 



264 DIFFERENTIAL EQUATIONS 

can be performed. This is the principle of energy in its simplest form. 
If two of the equations are multiplied by the chief variable of the other 
and subtracted, the result is 

yx" - xif = yX - xY (12) 

and the expression on the left is again an exact differential; if the 
right-hand side reduces to a constant or a function of t, then 



c' -xij' = I 



yx' -xij'= f(t) + C (12') 



is an integral of the equations. This is the principle of moment of 
momentum. If the equations can be multiplied by constants as 

Ix" + my" + nz" = IX + m Y + nZ, (13) 

so that the expression on the right reduces to a function of t, an inte- 
gration may be performed. This is the principle of momentum. These 
three are the most commonly usable devices. 

As an example : Let a particle move in a plane subject to forces attracting it 
toward the axes by an amount proportional to the mass and to the distance from 
the axes ; discuss the motion. Here the equations of motion are merely 

d 2 x , d 2 y 7 d 2 x 7 d 2 y 

m — i- = — kmx, m — = — kmy or = — kx, — - = — ky. 

dt 2 ' dt 2 dt 2 ' dt 2 

d 2 x d 2 y 

Then x V V — : 

dt 2 y dt 2 

A1 d 2 x d 2 y dx dy ' 
Also y — - — x — ~ = and y x — = C. 

y dt 2 dt 2 dt dt 

In this case the two principles of energy and moment of momentum give two 
integrals and the equations are reduced to two of the first order. But as it happens, 
the original equations could be integrated directly as 

— dx=- kxdx, (—) 2 = -kx 2 + C 2 , dX = dt 

dt 2 \dt) VC' 2 - kx 2 



-k(xdx + ydy) and (^)\ (^) 2 = - (x 2 + y 2 ) + C . 



%dy=-kydy, (^%-ky 2 + K 2 , -^- = dL 

dt 2 \dtj Vli 2 - ky 2 

The constants C 2 and K 2 of integration have been written as squares because they 
are necessarily positive. The complete integration gives 

Vkx = C sin (Vkt + C ± ), Vky = K sin ( Vkt + K. 2 ) . 

As another example : A particle, attracted toward a point by a force equal to 
r/m 2 + h 2 /r 3 per unit mass, where m is the mass and h is the double areal velocity 
and r is the distance from the point, is projected perpendicularly to the radius vec- 
tor at the distance Vmh ; discuss the motion. In polar coordinates the equations 
of motion are 



in 



Idt 2 



1 /d0\2"| = B= _rnr_ m/^ m d L 2 <W\ = $ = 
r\dt) J m 2 r 3 ' r dt\ dt) 



MOEE THAN" T^YO VARIABLES 265 

The second integrates directly as r 2 d<f>/dt = h where the constant of integration h 
is twice the areal velocity. Now substitute in the first to eliminate 0. 



d 2 r h 2 _ r Ji 2 , d 2 r r /dr\ 2 

dt 2 ~r 3 ~~m 2 ~r 3 d£ 2 



h r dr\ 2 r 2 „ 
-= or ( — ) = + C. 

i 2 m 2 \dt m 2 



Now as the particle is projected perpendicularly to the radius, dr/dt = at the 
start when r =vmh. Hence the constant C is h/m. Then 

dr _. _ r 2 dcp ,. . VmJidr 
— ^ ===== = dt and = dt give == = d<p. 

\ m m 2 \ Jim 



Hence V^ x \-\ = * + C or I--l=<*±^. 

Now if it be assumed that <p = at the start when r = V ?n/j, we find C = 0. 

Hence r 2 = — is the orbit. 

1 + 2 

To find the relation between and the time, 

r 2 dd> = hdt or = dt or t = m tan _1 0. 

1 + 2 

if the time be taken as t = when = 0. Thus the orbit is found, the expression 
of as a function of the time is found, and the expression of r as a function of the 
time is obtainable. The problem is completely solved. It will be noted that the 
constants of integration have been determined after each integration by the initial 
conditions. This simplifies the subsequent integrations which might in fact be 
impossible in terms T)f elementary functions without this simplification. 

EXERCISES 

1. Integrate these equations : 

dx dy dz 



yz xz xy 


0) 


. . dx dy dz 

(y) — = — = —, 

xz yz xy 


(5) 


dx dy dz 
( °- y = 3=1 + **' 


(f) 


2. Integrate the equations : 


(*) 


dx dy _ dz 


(7) 


x 2 + y 2 2 xy xz + yz ' 


(5) dx _ dy _ dz 


(«) 


' y 3 x — 2 x 4 2 y i — x s y yz (x 3 — y 3 ) 


dx dy dz 


0») 


' x {y 2 - z 2 ) y (z 2 -x 2 ) z (x 2 - y 2 ) ' 


(0) dx - dy - dz -dt, 


(0 



y 2 x 2 x 2 y 2 z 2 
dx dy dz 



dx _ dy dz 



1 3?/ + 4z 22/ + 02 
dx d?/ dz 



bz — cy ex — az ay — bx 

dx dy dz 

y + z x + z x + y' 

dx dy dz 



x (y-z) y(z- x) z (x - y) 
dx — dy dz 



x(y 2 -z 2 ) y(z 2 + x 2 ) z(x 2 + y 2 ) 
dx _ dy _ dz _ ^ 
y-z x+y x+z 'y-z x+y +t x+z+t 



266 DIFFEBENTIAL EQUATIONS 

3. Show that the differential equations of the orthogonal trajectories (curves 
of the family of surfaces F(x, y,z) = C are dx\dy.dz = F' x \F' y : F' z . Find the curves 
which cut the following families of surfaces orthogonally : 

(a) a 2 x 2 + b 2 f + c 2 * 2 = C, (/3) xyz = C, (y) y* = Cxz, 

(5) y — x tan (z + C), (e) y = x tan Cz, (f) 2 = Cxy. 

4. Show that the solution of dx : dy : dz = X : Y : Z, where X, F, Z are linear 
expressions in x, ?/, 2, can always be found provided a certain cubic equation can 
be solved. 

5. Show that the solutions of the two equations 

where T, T 1? T 2 are functions of t, may be obtained by adding the equation as 

^ (x + fy) + XT(x + Zz/) = T % -+ IT 2 

after multiplying one by I, and by determining X as a root of 
X 2 - (a + &') X + ab' - a'b = 0. 

6. Solve: (a)t— + 2(x-y)=t, t— + x+5y = t 2 , 

CLZ &Z 

(/3) tdx = (t- 2x)dt, tdy = (tx + ty + 2x - t)dt, 

Idx mdy ndz dt 



— + T(ax + by) = T 1 , -£+ T(a'x + b'y) = T 2 , 



(7) 



mn (y — z) nl (z — x) Im (x — y) t 



7. A particle moves in vacuo in a vertical plane under the force of gravity alone. 
Integrate. Determine the constants if the particle starts from the origin with a 
velocity V and at an angle of a degrees with the horizontal and at the time t — 0. 

8. Same problem as in Ex. 7 except that the particle moves in a medium which 
resists proportionately to the velocity of the particle. 

9. A particle moves in a plane about a center of force which attracts proportion- 
ally to the distance from the center and to the mass of the particle. 

10. Same as Ex. 9 but with a repulsive force instead of an attracting force. 

11. A particle is projected parallel to a line toward which it is attracted with 
a force proportional to the distance from the line. 

12. Same as Ex. 11 except that the force is inversely proportional to the square 
of the distance and only the path of the particle is wanted. 

13. A particle is attracted toward a center by a force proportional to the square 
of the distance. Find the orbit. 

14. A particle is placed at a point which repels with a constant force under 
which the particle moves away to a distance a where it strikes a peg and is 
deflected off at a right angle with undiminished velocity. Find the orbit of the 
subsequent motion. 

15. Show that equations (7) may be written in the form drxF = 0. Find the 
condition on F or on X, Y, Z that the integral curves have orthogonal surfaces. 



MOKE THAN TWO VARIABLES 267 

113. Introduction to partial differential equations. An equation 
which contains a dependent variable, two or more independent varia- 
bles, and one or more partial derivatives of the dependent variable 
with respect to the independent variables is called a partial differential 
equation. The equation 

p&y,z)fa + Q>( x ry,z)fy^ R ( x >y>*)> ^ = ^ 5 q = fy' ^ 

is clearly a linear partial differential equation of the first order in one 
dependent and two independent variables. The discussion of this equa- 
tion preliminary to its integration may be carried on by means of the 
concept of planar elements, and the discussion will immediately suggest 
the method of integration. 

When any point (x , y , z ) of space is given, the coefficients P, Q, R 
in the equation take on definite values and the derivatives p and q 
are connected by a linear relation. Now any planar element through 
(x Qf y , « ) may be considered as specified by the two slopes p and q ; for 
it is an infinitesimal portion of the plane z — z Q = p (x — x Q ) + q (y — y ) 
in the neighborhood of the point. This plane contains the line or lineal 
element whose direction is 

dxidy.dz =P:Q:R, (15) 

because the substitution of P, Q, R for dx = x — x Qf dy = y — y , 
dz = z — z Q in the plane gives the original equation Pp + Qq = R. 
Hence it appears that the planar elements defined by (14), of which 
there are an infinity through each point of space, are so related that all 
which pass through a given point of space pass through a certain line 
through- that point, namely the line (15). 

Now the problem of integrating the equation (14) is that of grouping 
the planar elements which satisfy it into surfaces. As at each point 
they are already grouped in a certain way by the lineal elements through 
which they pass, it is first advisable to group these lineal elements into 
curves by integrating the simultaneous equations (15). The integrals 
of these equations are the curves defined by two families of surfaces 
F(x, y, z) — C x and G(x, y, z) = C.,. These curves are called the charac- 
teristic curves or merely the characteristics of the equation (14). Through 
each lineal element of these curves there pass an infinity of the planar ele- 
ments which satisfy (14). It is therefore clear that if these curves be in 
any wise grouped into surfaces, the planar elements of the surfaces must 
satisfy (14) ; for through each point of the surfaces will pass one of the 
curves, and the planar element of the surface at that point must there- 
fore pass through the lineal element of the curve and hence satisfy (14). 



268 DIFFEBENTIAL EQUATIONS 

To group the curves F(x, y, z) = C\, G(x, y, z) = C 2 which depend 
on two parameters C 1} C 2 into a surface, it is merely necessary to intro- 
duce some functional relation C^—f^C^) between the parameters so 
that when one of them, as C v is given, the other is determined, and 
thus a particular curve of the family is fixed by one parameter alone 
and will sweep out a surface as the parameter varies. Hence to integrate 
(14), first integrate (15) and then write 

G(x, y, z) = *[F(x, y, *)] or *(F, G) = 0, (16) 

where $ denotes any arbitrary function. This will be the integral of 
(14) and will contain an arbitrary function <£. 

As an example, integrate (y — z)p + (z — x)q = x — y. Here the equations 

— = ^ = — give ^ + 2/ 2 + , 2 2_ x+y + z = O a 

y — z z — xx — y 

as the two integrals. Hence the solution of the given equation is 

x + y + z = $ (x 2 + y 2 + z 2 ) or <J> (x 2 + y 2 + z 2 , x + y + z) = 0, 

where $ denotes an arbitrary function. The arbitrary function allows a solution 
to be determined which shall pass through any desired curve ; for if the curve be 
/(x, y, z) = 0, g (x, y, z) = 0, the elimination of x, y, z from the four simultaneous 
equations 

F(x, y, z) = C lf G (x, y, z)=C 2 , f(x, y, z) = 0, g (x, y,z) = 

will express the condition that the four surfaces meet in a point, that is, that the 
curve given by the first two will cut that given by the second two ; and this elimi- 
nation will determine a relation between the two parameters C 1 and C 2 which will 
be precisely the relation to express the fact that the integral curves cut the given 
curve and that consequently the surface of integral curves passes through the given 
curve. Thus in the particular case here considered, suppose the solution were to 
pass through the curve y = x 2 , z = x ; then 

C y = x 2 , z = x 





x 2 + y 2 + z 2 = Cj, x + y + z 


give 


2x 2 + x 4 = C 1 , 


whence 


( Cf + 2C 2 -C 1 ) 2 +8C 2 - 



The substitution of C 1 = x 2 + y 2 + z 2 and C 2 = x + y + z in this equation will 
give the solution of (y — z)p+ (z — x)q—x — y which passes through the parabola 



114. It will be recalled that the integral of an ordinary differ- 
ential equation f(x, y, y', • ■ ■ , y {n) ) = of the nth order contains n con- 
stants, and that conversely if a system of curves in the plane, say 
F(x, y, C v •••, C n )= 0, contains n constants, the constants may be 
eliminated from the equation and its first n derivatives with respect 
to x. It has now been seen that the integral of a certain partial 
differential equation contains an arbitrary function, and it might be 



MOKE THAN TWO VAEIABLES 269 

inferred that the elimination of an arbitrary function would give 
rise to a partial differential equation of the first order. To show 
this, suppose F(x, y, z) = ®[G(x, y, z)~]. Then 

K + Kv = *' ■ (p* +-g',P), K + Ki = *' ■ (Pi + g'.v) 

follow from partial differentiation with respect to x and y ; and 

(F' z G' y - f;G' z ) P + (KG' a - F' Z G' X ) q = F;G' X - F' x G' y 

is a partial differential equation arising from the elimination of <£'. 
More generally, the elimination of n arbitrary functions will give rise 
to an equation of the nth. order; conversely it may be believed that 
the integration of such an equation would introduce n arbitrary func- 
tions in the general solution. 

As an example, eliminate from z = <£ (xy) + SP - (x + y) the two arbitrary func- 
tions <£ and <&. The first differentiation gives 

p = <£'.?/ + S^', q = $' • x + ty\ p — q = (y — x)&. 

Now differentiate again and let r — — -,s = — — , t = — - ■ Then 

ex 2 dxdy dy 2 

r — s = — #' + (y — x) *" • y, s — t = & + (y — x) <£" • x. 

These two equations with p — q = (y — x) <£' make three from which 

, . . . , . x + y. . d 2 z , , d 2 z t c 2 z x + y/dz dz\ 

xr-(x + y)s + yt= (p-q) or x —-{x + y)-—- + y— = r~~~~) 

x — y dx 2 cxdy dy 2 x — y\cx cy/ 

may be obtained as a partial differential equation of the second order free from 
$ and ^. The general integral of this equation would be z = <£ {xy) + & (x + y) . 

A partial differential equation may represent a certain definite type 
of surface. For instance by definition a conoidal surface is a surface 
generated by a line which moves parallel to a given plane, the director 
plane, and cuts a given line, the directrix. If the director plane be taken 
as z = and the directrix be the s-axis, the equations of any line of 
the surface are 

z = C v y = C 2 x, with C 1= ,<D(C 2 ) 

as the relation which picks out a definite family of the lines to form a 
particular conoidal surface. Hence z = 3> (y/x) may be regarded as the 
general equation of a conoidal surface of which z = is the director 
plane and the s-axis the directrix. The elimination of <E> gives px -\-qy=0 
as the differential equation of any such conoidal surface. 

Partial differentiation may be used not only to eliminate arbitrary func- 
tions, but to eliminate constants. For if an equation f(x, y, z, C v C 2 ) = 
contained two constants, the equation and its first derivatives with respect 
to x and y would yield three equations from which the constants could 



270 DIFFERENTIAL EQUATIONS 

be eliminated, leaving a partial differential equation F(x, y, z, p, q) = 
of the first order. If there had been five constants, the equation with 
its two first derivatives and its three second derivatives with respect 
to x and y would give a set of six equations from which the constants 
could be eliminated, leaving a differential equation of the second order. 
And so on. As the differential equation is obtained by- eliminating the 
constants, the original equation will be a solution of the resulting dif- 
ferential equation. 

For example, eliminate from z = Ax 2 + 2 Bxy + Cy 2 + Dx + Ey the five con- 
stants. The two first and three second derivatives are 

p = 2 Ax + 2 By + A q = 2 Bx + 2 Cy + E, r = 2 A, s = 2 B, t = 2C, 

Hence z = — ^ rx 2 — \ ty 2 — sxy + px + qy 

is the differential equation of the family of surfaces. The family of surfaces do 
not constitute the general solution of the equation, for that would contain two 
arbitrary functions, but they give what is called a complete solution. If there had 
been only three or four constants, the elimination would have led to a differential 
equation of the second order which need have contained only one or two of the 
second derivatives instead-of all three ; it would also have been possible to find three 
or two simultaneous partial differential equations by differentiating in cliff erent ways. 

115. If /(*, y, z, C\, C 2 ) = and F(x, y, z, p, q) = (17) 

are two equations of which the second is obtained by the elimination of 
the two constants from the first, the first is said to be the complete solu- 
tion of the second. That is, any equation which contains two distinct 
arbitrary constants and which satisfies a partial differential equation of 
the first order is said to be a complete solution of the differential equa- 
tion. A complete solution has an interesting geometric interpretation. 
The differential equation F = defines a series of planar elements 
through each point of space. So does f(x, y, z, C v C 2 ) = 0. For the 
tangent plane is given by 



dx 



(*-*.)+! 



(*-*o) = ° 

) 

with /(x , % , z , c v cy = 



(v-%)+ d £ 



as the condition that C 1 and C 2 shall be so related that the surface 
passes through (x , y Q , z ). As there is only this one relation between 
the two arbitrary constants, there is a whole series of planar elements 
through the point. As f(x, y, z, C v C 2 ) = satisfies the differential equa- 
tion, the planar elements defined by it are those defined by the differen- 
tial equation. Thus a complete solution establishes an arrangement of 
the planar elements defined by the differential equation upon a family 
of surfaces dependent upon two arbitrary constants of integration. 



MORE THAN TWO VARIABLES 271 

From the idea of a solution of a partial differential equation of the 
first order as a surface pieced together from planar elements which 
satisfy the equation, it appears that the envelope (p. 140) of any family 
of solutions will itself be a solution ; for each point of the envelope is 
a point of tangency with some one of the solutions of the family, and 
the planar element of the envelope at that point is identical with the 
planar element of the solution and hence satisfies the differential equa- 
tion. This observation allows the general solution to be determined from 
any complete solution. For if in f(x, y, z, C v C 2 ) = any relation 
C t) = &(C^) is introduced between the two arbitrary constants, there 
arises a family depending on one parameter, and the envelope of the 
family is found by eliminating C 1 from the three equations 

cf d® cf 

c, = Hc^ tc+d^tcr ' 1 /=a (18) 

As the relation C 2 = $(C 2 ) contains an arbitrary function 3>, the result 
of the elimination may be considered as containing an arbitrary func- 
tion even though it is generally impossible to carry out the elimination 
except in the case where 3> has been assigned and is therefore no longer 
arbitrary. 

A family of surfaces f(x, y, z, C v C ) = depending on two param- 
eters may also have an envelope (p. 139). This is found by eliminat- 
ing C 1 and C 2 from the three equations 

f(x, y, z, C v C 2 ) = 0, & = 0, |£ = 0. 

This surface is tangent to all the surfaces in the complete solution. 
This envelope is called the singular solution of the partial differential 
equation. As in the case of ordinary differential equations (§ 101), the 
singular solution may be obtained directly from the equation ; * it is 
merely necessary to eliminate p and a from the three equations 

F(x,y,z,p, q ) = 0, ^=0, ^=0. 

The last two equations express the fact that. F(p, q) = regarded as 
a function of p and q should have a double point (§ 57). A reference 
to § 67 will bring out another point, namely, that not only are all the 
surfaces represented by the complete solution tangent to the singular 
solution, but so is any surface which is represented by the general 
solution. 

* It is hardly necessary to point out the fact that, as in the case of ordinary equations, 
extraneous factors may arise in the elimination, whether of C\, C 2 or of p, q. 



272 DIFFEKENTIAL EQUATIONS 

EXERCISES 

1. Integrate these linear equations: 

{a) xzp + yzq = xy, (p) a (p + q) = z, (7) x 2 p + y 2 q = z 2 , 

(d) - yp + xq + 1 + z 2 = 0, (e) yp - xq = x 2 - y 2 , (f) (x + z)p = y, 
(77) x 2 p — xyq + 2/ 2 = 0, (0) (a — x)p + (6 — y) q = c — 2, 

( t ) p tan x + <Z tan ?/ = tan z, (k) (y 2 + z 2 — x 2 ) p — 2 x?/g + 2 xz = 0. 

2. Determine the integrals of the preceding equations to pass through the curves : 

for (a) x 2 + y 2 = l,z = 0, for (p) y = 0,x = z, 

for (7) y = 2x, z = 1, for (e) x = 2, y = 2. 

3. Show analytically that if _F(x, 2/, z) = C 1 is a solution of (15), it is a solution 
of (14) . State precisely what is meant by a solution of a partial differential equa- 
tion, that is, by the statement that F(x, y, z) = C x satisfies the equation. Show that 
the equations 

-. dz , _ dz _ , ^dF ^dF ^dF 
P — + Q — = R and P h Q }- R ■ — = 

dx dy - dx dy dz 

are equivalent and state what this means. Show that if F= C 1 and G = C 2 are 
two solutions, then F = $ (G) is a solution, and show conversely that a functional 
relation must exist between any two solutions (see § 62). 

4. Generalize the work in the text along the analytic lines of Ex. 3 to estab- 
lish the rules for integrating a linear equation in one dependent and four or n 
independent variables. In particular show that the integral of 

t> d z . , r. 8z td i ^ dx r dx n dz 

P 1 — + --- + P n — = Pn+i depends on —1 = • * • = — = , 

dx x cx n P x P n P n+1 

and that if F x — C v • • • , F n = C n are n integrals of the simultaneous system, the 
integral of the partial differential equation is ^(F^ • • •, F n ) = 0. 

_ T . . du du du 

5. Integrate : (a) x \- y \- z — = xyz, 

dx dy dz 

(|8) (y + z + u) ~ + (z + u + x) — + (u + x + y) — = x + y + z. 
dx cy dz 

6. Interpret the general equation of the first order F(x, y, z, p,q) = as deter- 
mining at each point (x , y , z ) of space a series of planar elements tangent to a 
certain cone, namely, the cone found by eliminating p and q from the three simul- 
taneous equations 

F(x Q , y Qi z , p, q). = 0, (x - x )p + {y-y )q = z- z , 

(x-x )-- { y-y )-=0. 

7. Eliminate the arbitrary functions: 

(a) x + y + z = $ (x 2 + y 2 + z 2 ), (/3) <i> (x 2 + y 2 , z - xy) = 0, 

(7) z = fr(x'+.y) + *(x-y), (5) z = e"v*(x-y), 

(e) 2 ; = 1 / 2 + 2<J>(x-i + log ? y), (fl $(-, V -, -) = 0. 

\?y z xl 



MOEE THAN TWO YAEIABLES 273 

8. Find the differential equations of these types of surfaces : 

(a) cylinders with generators parallel to the line x — az, y = bz, 

(/3) conical surfaces with vertex at (a, 6, c), 

(7) surfaces of revolution about the line x :y :z = a :b :c. 

9. Eliminate the constants from these equations: 

(a) z = (x + a)(y + b), (/3) a(x 2 + y 2 ) + bz 2 = 1, 

(7) (x - a) 2 + (y - bf + (z - c) 2 = 1, (5) (x - a) 2 + (y - b) 2 + (z - c) 2 = d 2 , 
(e) Ax 2 + 5X?/ + CV 2 + Dxz + Eyz = z 2 . 

10. Show geometrically and analytically that F(x, y, z) + aG(x, y, z) = b is a 
complete solution of the linear equation. 

11. How many constants occur in the complete solution of the equation of the 
third, fourth, or nth order ? 

12. Discuss the complete, general, and singular solutions of an equation of the 
first order F(x, y, z, u, u' x , u y , u z ) = with three independent variables. 

13. Show that the planes z = ax + by 4- C, where a and b are connected by the 
relation F(a, b) = 0, are complete solutions of the equation F(p, q) = 0. Integrate : 

(a) pq = 1, (/3) $ = p 2 + 1, (7) P 2 + q 2 = m 2 , 

(5) pq = k, (e) klogq+p = 0, (f) 3p 2 -2g 2 = 4pg, 

and determine also the singular solutions. 

14. Note that a simple change of variable will often reduce an equation to the 
type of Ex. 13. Thus the equations 



'& 



?)=0, F(xp,q) = 0, Wf- f) = 



with z = e g/ , x = e x/ , z = e 2 ', x = e x/ , y = ev', 

take a simpler form. Integrate and determine the singular solutions : 

(a) q = z+ px, (p) x 2 p 2 + y 2 q 2 = z 2 , (7) z = pq, 

(5) q = 2yp 2 , (e) (p - y) 2 + (q - x) 2 = 1, (f) 2 =p™g™. 

15. What is the obvious complete solution of the extended Clairaut equation 
z = xp -f yq + /(p, g) ? Discuss the singular solution. Integrate the equations : 
(a) z = xp+yq + Vp 2 + q 2 + 1, (j8) z = xp + ?/g + (p +_g) 2 , 
(7) z = xp + s/2 + pq, (5) z=xp + yq-2 Vpq. 

116. Types of partial differential equations. In addition to the 
linear equation and the types of Exs. 13-15 above, there are several 
types which should be mentioned. Of these the first is the general 
equation of the first order. If F(x, y, z, p, (f) = is the given equation 
and if a second equation <3> (x, y, z, p, q, a) = 0, which holds simultane- 
ously with the first and contains an arbitrary constant can be found, 
the two equations may be solved together for the values of p and q, and 
the results may be substituted in the relation dz = pdx -f- qdy to give a 
total differential equation of which the integral will contain the con- 
stant a and a second constant of integration b. This integral will then 



274 



DIFFERENTIAL EQUATIONS 



be a complete integral of the given equation ; the general integral may 
then be obtained by (18) of § 115. This is known as Charpit's method. 
To find a relation <£ = differentiate the two equations 

F(x, y, z, p, q) = 0, $(x, y, z, p, q, a) = 

with respect to x and y and use the relation that dz be exact. 



(19) 



x ^ zl ^ p dx^ q dx ' 


*;> 


x ^ zl ^ p dx ^ q dx > 


-■«?. 


y ^ ^^ p dy* q dy ' 


*;» 


a/ _i_ & a _i_ <f>' JP. _i_ 3/ _£ — 


-*;. . 







Multiply by the quantities on the right and add. Then 

c& , , d& c$? d$> , d$> 

(F<+pF;)-^ + (F;+ q F;)^-F;^-F' q ^-(pF;,+ q F;) ^ = 0. (20) 

Now this is a linear equation for <X> and is equivalent to 

dp dq dx dy dz d$? 



K+pK K + 9.K -K -F' q -(p F ; + q F£ o (21) 

Any integral of this system containing p or q and a will do for <!>, and 
the simplest integral will naturally be chosen. 

As an example take zp(x + y) + p(q — p) — z 1 = 0. Then Charpit's equa- 
tions are 

dp _ dq _ dx 

— zp+p 2 (x + y) zp — 2zq + pq(x + y) 2p — q — z(x + y) 

_ dy__ dz 

— p 2p 2 —2pq—pz(x+y) 

How to combine these so as to get a solution is not very clear. Suppose the sub- 
stitution z = e z ', p = e z 'p', q — e z 'q' be made in the equation. Then 

p' {x + y) + p' {q' — p') — 1 = 
is the new equation. Tor this Charpit's simultaneous system is 
dp' _ dq' _ dx _ dy _ dz 

p' ~ p' ~ 2p' — q' - (x ■ + y) —p' 2p' 2 — 2pq — p'(x + y) 

The first two equations give at once the solution dp' = dq' or q' = p' + a. Solving 
p' (x + y) + p' (q' — p') — 1 = and q' = p' + a, 

1 * ', 1 



V 



a + x + 



a + x + y 



+ a, 



, , dx + dy~ 

dz' — h 

a + x + y 



MOKE THAN TWO VARIABLES 275 

Then z = log {a + x + y) + ay + b or log z = log (a + x + y) + ay + b 

is a complete solution of the given equation. This will determine the general 
integral by eliminating a between the three equations 

z = e a y + b (a + x + y), b=f(a), = (y +f'(a))(a + x + y) + 1, 

where f(a) denotes an arbitrary function. The rules for determining the singular 
solution give z — ; but it is clear that the surfaces in the complete solution can- 
not be tangent to the plane z = and hence the result 2 = must be not a singular 
solution but an extraneous factor. There is no singular solution. 

The method of solving a partial differential equation of higher order 
than the first is to reduce it first to an equation of the first order and 
then to complete the integration. Frequently the form of the equation 
will suggest some method easily applied. For instance, if the deriva- 
tives of lower order corresponding to one of the independent variables 
are absent, an integration may be performed as if the equation were 
an ordinary equation with that variable constant, and the constant of 
integration may be taken as a function of that variable. Sometimes a 
change of variable or an interchange of one of the independent variables 
with the dependent variable will simplify the equation. In general the 
solver is left mainly to his own devices. Two special methods will be 
mentioned below. 

117. If the equation is linear with constant coefficients and all the 
derivatives are of the same order, the equation is 

(a D£ + a x D* ~^D y + • • • + ** _ A*>; ~ l + a, n D%) z = R(x,y). (22) 

Methods like those of § 95 may be applied. Factor the equation. 

a (D x - aj) y ) (D x - a 2 D y ) • • • (D x - a n D y ) z = R (x, y). (22') 

Then the equation is reduced to a succession of equations 

Zys - aD y z = R (x, y), 

each of which is linear of the first order (and with constant coefficients). 
Short cuts analogous to those previously given may be developed, but 
will not be given. If the derivatives are not all of the same order but 
the polynomial can be factored into linear factors, the same method will 
apply. For those interested, the several exercises given below will serve 
as a synopsis for dealing with these types of equation. 
There is one equation of the second order,* namely 

J_ Pu _ c^u chi dhc 

f 2 df ~ dx' + of + a* 2 ' ^ l6) 

* This is one of the important differential equations of physics ; other important equa- 
tions and methods of treating them are discussed in Chap. XX. 



276 DIFEEBENTIAL EQUATIONS 

which occurs constantly in the discussion of waves and which has there- 
fore the name of the wave equation. The solution may be written down 
by inspection. For try the form 

u (x, y, z, t) = F(ax + by + cz — Vt) + G(ax + by + cz + Vt). (24) 

Substitution in the equation shows that this is a solution if the relation 
a 2 + b' 2 + <r = 1 holds, no matter what functions F and G may be. Note 
that the equation 

ax + by + cz — Vt = 0, a 2 + b' 2 -f c 2 = 1, 

is the equation of a plane at a perpendicular distance Vt from the origin 
along the direction whose cosines are a, b, c. If t denotes the time and 
if the plane moves away from the origin with a velocity V, the function 
F(ax + by + cz — Vt) = F(0) remains constant ; and if G = 0, the value 
of u will remain constant. Thus u = F represents a phenomenon which 
is constant over a plane and retreats with a velocity V, that is, a plane 
wave. In a similar manner u— G represents a plane wave approaching 
the origin. The general solution of (23) therefore represents the super- 
position of an advancing and a retreating plane wave. 

To Monge is due a method sometimes useful in treating differential equations 
of the second order linear in the derivatives r, s, t ; it is known as Mongers method. 

Let Rr + Ss + Tt = V (25) 

be the equation, where R, S, T, V are functions of the variables and the derivatives 
p and q. From the given equation and 

dp = rdx + sdy, dq = sdx + tdy, 

the elimination of r and t gives the equation 

s (Edy 2 - Sdxdy + Tdx 2 ) - (Rdydp + Tdxdq - Vdxdy) = 0, 

and this will surely be satisfied if the two equations 

Rdy 2 - Sdxdy + Tdx 2 = 0, Rdydp + Tdxdq - Vdxdy = (25') 

can be satisfied simultaneously. The first may be factored as 

dy — /j (x, y, z, p, q) dx = 0, dy — f. 2 (x, y, z, p, q) dx = 0. (26) 

The problem then is reduced to integrating the system consisting of one of these fac- 
tors with (25') and dz =pdx + qdy, that is, a system of three total differential equations. 
If two independent solutions of this system can be found, as 

u x (x, y, z, p, q) = Ci, u 2 (x, y, z, p, q) = C 2 , 

then u x — <£ (u 2 ) is a first or intermediary integral of the given equation, the general 
integral of which may be found by integrating this equation of the first order. If 
the two factors are distinct, it may happen that the two systems which arise may 
both be integrated. Then two first integrals u x = $ (m 2 ) and v 1 - < ^ (v 2 ) will be found, 
and instead of integrating one of these equations it may be better to solve both for 
p and q and to substitute in the expression dz = pdx + qdy and integrate. When, 
however, it is not possible to find even one first integral, Monge's method fails. 



MORE THAN TWO VARIABLES 277 

As an example take (x + y) (r — t) = — 4p. The equations are 

(x + y) dy 2 — (x + y) dx 2 = or dy — dx = 0, dy + dx = 
and (x + y) dydp — (x + y) dxdq + ipdxdy = 0. (A) 

Now the equation dy — dx = may be integrated at once to give y = x + C x . The 
second equation (A) then takes the form 

2 xdp + 4 pdx — 2xdq + C l {dp — dq) = ; 
but as dz = pdx + qdy = (p -\- q)dx in this case, we have by combination 

2 (xdp + pdx) — 2 {xdq + gdx) + C'j (dp — dq) + 2dz = 
or (2x+C 1 )Q9- 5 ) + 22 = C, or (x + y) (p - q) + 2z = C 2 . 

Hence (x + y) (p - g) + 2 2 = <J> (y - x) (27) 

is a first integral. This is linear and may be integrated by 

dx dy dz ^ dx dz 

or x + y 



x + y x + y <i> (2/ - x) - 2 z K x &(K x -2x)-2z 

This equation is an ordinary linear equation in z and x. The integration gives 

2.r n 2s" 

2 x) dx + K 2 , 



: I eKi$(E l - 

2x p ' 



2x s% 2x 

Hence (x + y) ze x + v— I e K i$(K x — 2x)dx = K 2 = <k (K x ) = <k(x + ij) 



is the general integral of the given equation when K x has been replaced by x + y 
after integration, — an integration which cannot be performed until <l> is given. 

The other method of solution would be to use also the second system containing 
dy + dx = instead of dy — dx = 0. Thus in addition to the first integral (27) a 
second intermediary integral might be sought. The substitution of dy + dx = 0, 
y + x = C x in (A) gives Cj (dp + dq) + 4pdx = 0. This equation is not integrable, 
because dp + dq is a perfect differential and pdx is not. The combination with 
dz = pdx + qdy = (p — q)dx does not improve matters. Hence it is impossible to 
determine a second intermediary integral, and the method of completing the 
solution by integrating (27) is the only available method. 

Take the equation ps — qr — 0. Here S = p, R =— q, T = V — 0. Then 

— qdy 2 — pdxdy = or dy = 0, pdx + qdy — and — qdydp = 

are the equations to work with. The system dy = 0, qdydp = 0, dz = pdx + qdy, 
and the system pdx + qdy = 0, qdydp = 0, dz = pdx. + gdy are not very satisfactory 
for obtaining an intermediary integral u x — <£ (u 2 ). although p = <£> (z) is an obvious 
solution of the first set. It is better to use a method adapted to this special 
equation. Note that 

a /,\ = !*_£ _ and a /,\ = , = 

ex \p/ p 2 cx \pj p 

By (11), p. 124, 1 = - (p) ; then ^ = - f(y) 

P \cyJz cy 

and x =- J/(y) ^ + * (z) = $ (y) + * (z). 



278 DIFFERENTIAL EQUATIONS 

EXERCISES 

1. Integrate these equations and discuss the singular solution: 

(a) p$+q* = 2x, (j8) {p 2 + q 2 )x = pz, (7) (p + q) (px + qy) = 1, 

(5) pq =px + qy, (e) p 2 + q 2 =x + y, (?) xp 2 - 2zp + xy = 0, 

( V ) q* = z 2 (p-q), (0) q(p 2 z + q 2 ) = l, (<) p (1 + q 2 ) = q (z - c), 

( k ) ap (1 + g) = qz, (X) y 2 (p 2 - 1) = x 2 p 2 , ( M ) z 2 (p 2 + g 2 + 1) = c 2 , 

(y) p = (z + ?/g) 2 , (0) pz = 1 + g 2 , (tt) 2 - pg = 0, (p) q = xp + p 2 . 

2. Show that the rule for the type of Ex. 13, p. 273, can be deduced by Charpit's 
method. How about the generalized Clairaut form of Ex. 15 ? 

3. (a) Eor the solution of the type /^x, p) = f 2 (y, g), the rule is : Set 

f 1 (x,P)=f 2 (y, ?) = «, 
and solve for p and g asp = g 2 (x, a), g = g 2 (y, a) ; the complete solution is 

z = J g^x, a)dx + J g 2 (y, a)dy + 6. 

(/3) For the type .F(z, p, q) = the rule is : Set X = x + ay, solve 

F ( z ^' a S for ^=* <z - a) ' andiet /^= /<2 ' a); 

the complete solution is x -f ay + 6 = /(z, a). Discuss these rules in the light of 
Charpit's method. Establish a rule for the type F(x + y, p, q) = 0. Is there any 
advantage in using the rules over the use of the general method ? Assort the exam- 
ples of Ex. 1 according to these rules as far as possible. 

4. What is obtainable for partial differential equations out of any characteristics 
of homogeneity that may be present ? 

5. By differentiating p = /(x, y, z, q) successively with respect to x and y show 
that the expansion of the solution by Taylor's Formula about the point (x , y , z ) 
may be found if the successive derivatives with respect to y alone, 

cz C 2 Z c s z c n z 

cy cy 2 cy 8 cy n 

are assigned arbitrary values at that point. Note that this arbitrariness allows the 
solution to be passed through any curve through (x , y , z ) in the plane x = x . 

6. Show that F(x, y, z, p, q) = satisfies Charpit's equations 

dx dy dz dp dg 

du = = = = = : > UO) 

-K -K -(pK + vK) K + pK K + *K 

where u is an auxiliary variable introduced for symmetry. Show that the first 
three equations are the differential equations of the lineal elements of the cones of 
Ex. 6, p. 272. The integrals of (28) therefore define a system of curves which have 
a planar element of the equation F = passing through each of their lineal tan- 
gential elements. If the equations be integrated and the results be solved for the 
variables, and if the constants be so determined as to specify one particular curve 
with the initial conditions x , y , z , p , q , then 

x = x(u, x , y , z ,p , g ), y = y (•••), « = «(■ • •), P--P (•'•), Q = «(•■■)• 



MORE THAN TWO VARIABLES 279 

Note that, along the curve, q =f(p) and that consequently the planar elements 
just mentioned must lie upon a developable surface containing the curve (§ 67). The 
curve and the planar elements along it are called a characteristic and a characteristic 
strip of the given differential equation. In the case of the linear equation the 
characteristic curves afforded the integration and any planar element through 
their lineal tangential elements satisfied the equation ; but here it is only those 
planar elements which constitute the characteristic strip that satisfy the equation. 
What the complete integral does is to piece the characteristic strips into a family 
of surfaces dependent on two parameters. 

7. By simple devices integrate the equations. Check the answers: 

(") % = /(*), (0) ^ = 0, (7) ^- = - + a, 

ex 2 cy n dxdy y 

(8) s + pf(x) = g\y), (e) ar = xy, (f) xr = (n-l)p. 

8. Integrate these equations by the method of factoring: 

(a) (D 2 - o»Dj) z = 0, 08) (D« - B y f z = 0, ( 7 ) (D x D 2 y -!>»)* = 0, 
(5) (2^+32)^+ 22^)« = jj + y, (e) (BJ - D.A, - 6 D 2 ) z = xy, 
(f) (l^-I^-3D x + 3J) y ) z = 0, (*) (D 2 _D 2 + 2D x +l)z = e -*. 

9. Prove the operational equations : 

(a) e«**>„0 (y) = (1 + aarD,, + h a 2 x 2 B 2 y + . . .) fy) = <f> fy + ax), 

(/3) — — = e«j>y — = e«*A, <f>(y) = <f>(y+ ax), 

(7) -=— — — B(x, y) = e"^„ f V«*^fi(|, y)d£ = f "fl (£, y + aa; - arf ) <*£• 
U x — al)y J J 

10. Prove that if [(D x - ^A,)™! • ■ • (D x - a*Dy)"*] z = 0, then 

2 = <i> u (?/ + ar^) + !B* la (y + ar^x) + • • • + a;"*- 1 .*!*,^ + a t x) + • • • 

+ **i(» + W) + aB*M(y + a&) + • • • + x^k-^Skm^y + ar^), 
where the <£'s are all arbitrary functions. This gives the solution of the reduced equa- 
tion in the simplest case. What terms would correspond to (D x — aD y — fi) m z = ? 

11. Write the solutions of the equations (or equations reduced) of Ex. 8. 

12. State the rule of Ex. 9 (7) as : Integrate R(x, y — ax) with respect to x and 
in the result change y to y + ax. Apply this to obtaining particular solutions of 
Ex.8 (5), (e), (rj) with the aid of any short cuts that are analogous to those of 
Chap. VIII. 

13. Integrate the following equations : 

(a) (Bl -D 2 xy + D y -l)z = cos (x + 2 y) + ev, (/3) x 2 r 2 + 2 xys + y 2 t 2 = x 2 + y 2 , 

(7) (D x + D xy + Dy-l)z = sin(x + 2y), (8) r - t- Sp + Sq = e* + *v, 

CO M-2^ + D;) 2 = x-2 (f) r -<+i> + 3g-2* = e*- v-x 2 y, 

(97) (Dl-D x D y -2Dl + 2D X + 2D y ) 2 = e 2*+«y+ sin(2x + y) + xy. 

14. Try Monge's method on these equations of the second order : 

(a) q 2 r - 2pqs + p 2 t = 0. (/3) r - a 2 t = 0, (7) r + s .= - p, 

(5) ff(l + g)r— (p + g + 2i)g)»+p(l+p)*=0, (e) x 2 r + 2 xys + yH = 0, 

( t ) (& + cg) 2 r - 2 (6 + eg) (a + cp) s + (a + cp) 2 £ = 0, ( v ) r + ka 2 t = 2 as. 
If any simpler method is available, state what it is and apply it also. 



280 DIFFERENTIAL EQUATIONS 

15. Show that an equation of the form Rr + Ss + Tt + U (rt — s 2 ) = V neces- 
sarily arises from the elimination of the arbitrary function from 

Mx, y, z , p 5 q) =fl u 2 ( x i y-< *■> p, s)L 

Note that only such an equation can have an intermediary integral. 

16. Treat the more general equation of Ex. 15 by the methods of the text and 
thus show that an intermediary integral may be sought by solving one of the systems 

Udy + \ Tdx + \ Udp = 0, Udx + \Bdy + \ Udq = 0, 

Udx + \Bdy + \Udq = 0, Udy + \Tdx + \Udp = 0, 

dz — pdx + qdy, dz = pdx + qdy, 

where \ and X 2 are roots of the equation \ 2 (RT + UV) + \US + 77 2 = 0. 

17. Solve the equations : (a) s 2 — rt dt 0, (£) s 2 — ri = a 2 , 

(7) ar + 6s + c£ + e (rt — s 2 ) = h, (5) xqr + ypi + xy (s 2 — r£) = pq. 



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